hdu

http://acm.hdu.edu.cn/showproblem.php?pid=1172

这个题换一种想法,可以找出四位数中所有满足条件的数看是否只有一个.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#pragma comment(linker, "/STACK:102400000,102400000")
#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("b.txt", "w", stdout);
#define maxn 1000000000
#define N 2510
#define mod 1000000000
using namespace std;

struct point
{
    int x,y,z;
}p[105];

bool judge(int n,int m)
{
    int a[5],b[5],used[5];
    a[0]=n/1000,a[1]=n/100%10,a[2]=n/10%10,a[3]=n%10;
    b[0]=p[m].x/1000,b[1]=p[m].x/100%10,b[2]=p[m].x/10%10,b[3]=p[m].x%10;
   // for(int i=0;i<4;i++)
       // printf("%d ",b[i]);
        //printf("
");
    int ans=0;
    for(int i=0;i<4;i++)
    {
        if(a[i]==b[i]) ans++;
    }
   // printf("%d
",ans);
    if(ans!=p[m].z) return false;
    ans=0;
    memset(used,0,sizeof(used));
    for(int i=0;i<4;i++)
        for(int j=0;j<4;j++)
        if(!used[j]&&a[i]==b[j])
        {
             ans++;
             used[j]=1;
             //printf("%d %d
",a[i],b[j]);
             break;
        }
   // printf("%d
",ans);
    if(ans!=p[m].y) return false;
    return true;
}
int main()
{
   //freopen("a.txt","r",stdin);
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++) scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);
        int ans=0,cnt=0;
        for(int i=1000;i<=9999;i++)
        {
            bool flag=0;
            for(int j=0;j<n;j++)
                if(!judge(i,j)) {flag=1;break;}
            if(!flag) {ans++;cnt=i;}
        }
      //  printf("%d
",ans);
        if(ans==1) printf("%d
",cnt);
        else printf("Not sure
");
    }
    return 0;
}