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[10]={490,288,242,434,312,410,474,290,506,442};
int temp;
char  ch[50];
int a[4],b[4];

int main()
{
    int i,j,x,y,figure,group,Case=0,_i;
    //freopen("in.txt","r",stdin);
    scanf("%d",&group);gets(ch);
    while(group--){
        mst(a,0);mst(b,0);
        for(i=0; i<3; ++i){
            _i=0;
            gets(ch);
            for(j=0; j<12; ++j){
                if(ch[_i++]!=‘ ‘){
                    a[j/3]|=1<<(i*3+j%3);
                }
            }
            ++_i;
            for(j=0; j<12; ++j){
                if(ch[_i++]!=‘ ‘){
                    b[j/3]|=1<<(i*3+j%3);
                }
            }
        }
        int index=0;
        int ansh,ansm;
        for(i=0; i<1440; ++i){
            int j=(i-15+1440)%1440;
            int h1=i/60;
            int m1=i%60;
            int h2=j/60;
            int m2=j%60;
            if((num[h1/10]&a[0])!=a[0])
                continue;
            if(int(num[h1%10]&a[1])!=a[1])
                continue;
            if(int(num[m1/10]&a[2])!=a[2])
                continue;
            //cout<<1<<endl;
            if((num[m1%10]&a[3])!=a[3])
                continue;
            if((num[h2/10]&b[0])!=b[0])
                continue;
            if((num[h2%10]&b[1])!=b[1])
                continue;
            if((num[m2/10]&b[2])!=b[2])
                continue;
            if((num[m2%10]&b[3])!=b[3])
                continue;
            if(++index>1) break;
            ansh=h1;ansm=m1;
        }
        if(index>1) printf("Not Sure\n");
        else printf("%d%d%d%d\n",ansh/10,ansh%10,ansm/10,ansm%10);
    }
    return 0;
}

参考:传送门