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Merge pull request ashutosh97#17 from Jee-Van-Kur/new-prob
A new problem added
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‎The Reversed World/Solution.cpp

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#include <iostream>
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#include <algorithm>
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#include <string>
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#include <vector>
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#include <cmath>
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#include <cstdio>
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#include <queue>
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#include <list>
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#include <stack>
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#include <utility>
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#include <numeric>
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#include <map>
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#include <cctype>
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#include <cstring>
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#include <sstream>
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#include <cstdlib>
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#include <cassert>
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#include <iomanip>
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#include <set>
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using namespace std;
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#define F(a,b) for(int a=0;a<b;a++)
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#define REP(a,b) for(int a=0;a<b;a++)
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#define FOR(a,b,c) for(int a=b;a<c;a++)
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#define FORD(a,b,c) for(int a=b;a>=c;a--)
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#define S scanf
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#define P printf
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#define LEN(x) ((int)x.length())
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#define SZ(x) ((int)x.size())
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#define ALL(x) x.begin(), x.end()
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#define MP(x,y) make_pair(x,y)
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#define PB(x) push_back(x)
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#define INF 1000000000
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typedef long long LL;
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typedef pair<int,int> PII;
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typedef pair<int, PII> PIII;
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typedef vector<int> VI;
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typedef vector<bool> VB;
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typedef vector<string> VS;
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//int d[][2]={{-1.0},{1,0},{0,-1},{0,1}};
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char a[1000005];
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char b[1000005];
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char c[1000005];
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char minN[1000005], maxN[1000005];
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//char ans[1000005];
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//char D[1000005];
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int digC[10];
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int fDC[10];
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int tDC;
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int n;
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bool solveB(int st)
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{
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b[0] = 0;
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int tmpst = st;
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int i = n-1;
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VB db(n);
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bool fAns = false;
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while( i < n )
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{
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if( !db[i] )
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{
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db[i] = true;
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if( digC[ minN[i] - '0' ] )
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{
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a[i] = minN[i];
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digC[ minN[i] - '0' ]--;
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i--;
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}
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else
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{
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FORD( j, minN[i]-'0' - 1, 0 )
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{
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if( digC[j] )
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{
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a[i] = j + '0';
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//keep Digit count
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if( !fAns )
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{
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tDC = 0;
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REP(xx,10) { fDC[xx] = digC[xx]; tDC+=digC[xx]; }
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strcpy( b, a+i );
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fDC[j]--;
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tDC--;
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}
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else
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{
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int za = 0, zb = 0;
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bool fl = false;
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REP(xx,10)
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{
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za += fDC[xx];
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zb += digC[xx] + ((j==xx)?-1:0);
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if( za == tDC )
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{
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if( zb > za )
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{
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if( b[0]-'0' > xx ) fl =true;
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}
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break;
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}
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if( za < zb ) fl = true;
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else if( za > zb ) break;
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}
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if( fl )
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{
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tDC = 0;
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REP(xx,10) { fDC[xx] = digC[xx]; tDC+=digC[xx]; }
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strcpy( b, a+i );
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fDC[j]--;
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tDC--;
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}
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}
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fAns = true;
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}
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}
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i++;
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}
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}
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else
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{
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if( minN[i] == a[i] )
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digC[ a[i]-'0' ]++;
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FORD( j, minN[i]-'0' - 1, 0 )
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{
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if( digC[j] )
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{
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a[i] = j + '0';
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//keep Digit count
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if( !fAns )
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{
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tDC = 0;
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REP(xx,10) { fDC[xx] = digC[xx]; tDC+=digC[xx]; }
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strcpy( b, a+i );
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fDC[j]--;
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tDC--;
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}
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else
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{
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int za = 0, zb = 0;
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bool fl = false;
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REP(xx,10)
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{
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za += fDC[xx];
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zb += digC[xx] + ((j==xx)?-1:0);
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if( za == tDC )
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{
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if( zb > za )
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{
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if( b[0]-'0' > xx ) fl =true;
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}
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break;
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}
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if( za < zb ) fl = true;
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else if( za > zb ) break;
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}
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if( fl )
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{
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tDC = 0;
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REP(xx,10) { fDC[xx] = digC[xx]; tDC+=digC[xx]; }
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strcpy( b, a+i );
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fDC[j]--;
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tDC--;
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}
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}
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fAns = true;
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}
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}
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i++;
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}
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}
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if( fAns )
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{
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REP(xx,10)
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{
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while( fDC[xx] )
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{
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a[tmpst] = xx + '0';
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tmpst++;
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fDC[xx]--;
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}
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}
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a[tmpst] = 0;
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strcat(a,b );
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puts(a);
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return true;
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}
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return false;
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}
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bool solveF()
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{
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a[n] = 0;
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int i = 0;
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VB d(n);
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while( i >= 0 )
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{
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if( !d[i] )
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{
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d[i] = true;
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if( digC[ maxN[i]-'0' ] )
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{
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a[i] = maxN[i];
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digC[ maxN[i]-'0' ]--;
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i++;
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}
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else
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{
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FOR( j, maxN[i]-'0' + 1, 10 )
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{
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if( digC[j] )
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{
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a[i] = j + '0';
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digC[j]--;
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if( solveB(i+1) ) return true;
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digC[j]++;
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}
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}
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i--;
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}
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}
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else
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{
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if( maxN[i] == a[i] )
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digC[ a[i]-'0' ]++;
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FOR( j, maxN[i]-'0' + 1, 10 )
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{
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if( digC[j] )
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{
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a[i] = j + '0';
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digC[j]--;
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if( solveB(i+1) ) return true;
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digC[j]++;
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}
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}
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i--;
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}
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}
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return false;
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}
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int main()
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{
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int t; S("%d", &t );
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while( t-- )
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{
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S("%d", &n);
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gets(a);
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gets(a);
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gets(b);
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gets(c);
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//S("%s%s%s", a, b, c );
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if( strcmp(a,b) > 0 )
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strcpy( maxN, a );
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else
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strcpy( maxN, b );
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reverse(a, a + n );
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reverse(b, b + n );
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if( strcmp(a,b) < 0 )
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strcpy( minN, a );
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else
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strcpy( minN, b );
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REP(i,10) digC[i] = 0;
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REP(i,n ) digC[ c[i]-'0' ]++;
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int ax = 0, ay = n-1;
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REP(i,10)
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{
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REP(j,digC[i])
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{
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c[ax] = i+'0';
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a[ay] = i + '0';
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ax++; ay--;
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}
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}
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c[n] = 0; a[n] = 0;
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if( strcmp( a, maxN ) > 0 && strcmp( c, minN ) < 0 )
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{
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reverse( minN, minN + n );
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if( !solveF() ) P("-1\n");
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}
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else puts("-1");
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}
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return 0;
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}

‎The Reversed World/The Reversed World.txt

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Given three numbers A, B and C all having N digits, little Johnny needs to write a program to permute the digits of the number C so that the new number is greater than both A and B.
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This problem would be so easy if Johnny's teacher didn't require that Johnny's program must be correct in both the real world and the reversed world. What happens once the world is reversed? One thing Johnny knows for sure is that all the numbers are also reversed. The second thing, the "greater than" relation becomes the "smaller than" relation.
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For the sake of clarity, we will use aR to denote the number a in the reversed world (for instance, 113R = 311).
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Thus Johnny task is now permute the digits of C so that the new number is greater than both A and B. In additional, the reverse of the new number should be smaller than both AR and BR. If there are multiple possible solutions, Johnny needs to find the smallest one.
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As usual, please help Little Johnny with his task!
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#Input
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The first line contains a number T (about 10) which is the number of the test cases. Each test case has the following form.
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The first line contains N (1 <= N <= 1000000).
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The second line contains the number A.
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The third line contains the number B.
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The fourth line contains the number C.
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Each test case's input is separated by a blank line.
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Output
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For each test case, print the solution in a single line. If there is no solution, print the number -1.
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Example
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#Input:
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2
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2
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12
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23
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32
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3
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124
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324
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335
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#Output:
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-1
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353
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Time Limit 0.26s
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