poj1676（转换为二进制求解）

题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=13175

题目关键:将0~9十个数转换为二进制数进行枚举比较

int num[10]={490,288,242,434,312,410,474,290,506,442};

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <cstring>

#include <stack>

#include <cctype>

#include <queue>

#include <string>

#include <vector>

#include <set>

#include <map>

#include <climits>

#define lson root<<1,l,mid

#define rson root<<1|1,mid+1,r

#define fi first

#define se second

#define ping(x,y) ((x-y)*(x-y))

#define mst(x,y) memset(x,y,sizeof(x))

#define mcp(x,y) memcpy(x,y,sizeof(y))

#define Min(x,y) (x<y?x:y)

#define Max(x,y) (x>y?x:y)

using namespace std;

#define gamma 0.5772156649015328606065120 //欧拉常数

#define MOD 100000007

#define inf 0x3f3f3f3f

#define N 105

#define maxn 10001000

typedef long long LL;

typedef pair<int,int> PII;

int num[]={,,,,,,,,,};

int temp;

char  ch[];

int a[],b[];

int main()

{

    int i,j,x,y,figure,group,Case=,_i;

    //freopen("in.txt","r",stdin);

    scanf("%d",&group);gets(ch);

    while(group--){

        mst(a,);mst(b,);

        for(i=; i<; ++i){

            _i=;

            gets(ch);

            for(j=; j<; ++j){

                if(ch[_i++]!=' '){

                    a[j/]|=<<(i*+j%);

                }

            }

            ++_i;

            for(j=; j<; ++j){

                if(ch[_i++]!=' '){

                    b[j/]|=<<(i*+j%);

                }

            }

        }

        int index=;

        int ansh,ansm;

        for(i=; i<; ++i){

            int j=(i-+)%;

            int h1=i/;

            int m1=i%;

            int h2=j/;

            int m2=j%;

            if((num[h1/]&a[])!=a[])

                continue;

            if(int(num[h1%]&a[])!=a[])

                continue;

            if(int(num[m1/]&a[])!=a[])

                continue;

            //cout<<1<<endl;

            if((num[m1%]&a[])!=a[])

                continue;

            if((num[h2/]&b[])!=b[])

                continue;

            if((num[h2%]&b[])!=b[])

                continue;

            if((num[m2/]&b[])!=b[])

                continue;

            if((num[m2%]&b[])!=b[])

                continue;

            if(++index>) break;

            ansh=h1;ansm=m1;

        }

        if(index>) printf("Not Sure\n");

        else printf("%d%d%d%d\n",ansh/,ansh%,ansm/,ansm%);

    }

    return ;

}

参考:传送门