hdu - 1172 猜数字 (思维题)

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\n", 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[];
 
 bool judge(int n,int m)
 {
     int a[],b[],used[];
     a[]=n/,a[]=n/%,a[]=n/%,a[]=n%;
     b[]=p[m].x/,b[]=p[m].x/%,b[]=p[m].x/%,b[]=p[m].x%;
    // for(int i=0;i<4;i++)
        // printf("%d ",b[i]);
         //printf("\n");
     int ans=;
     for(int i=;i<;i++)
     {
         if(a[i]==b[i]) ans++;
     }
    // printf("%d\n",ans);
     if(ans!=p[m].z) return false;
     ans=;
     memset(used,,sizeof(used));
     for(int i=;i<;i++)
         for(int j=;j<;j++)
         if(!used[j]&&a[i]==b[j])
         {
              ans++;
              used[j]=;
              //printf("%d %d\n",a[i],b[j]);
              break;
         }
    // printf("%d\n",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=;i<n;i++) scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);
         int ans=,cnt=;
         for(int i=;i<=;i++)
         {
             bool flag=;
             for(int j=;j<n;j++)
                 if(!judge(i,j)) {flag=;break;}
             if(!flag) {ans++;cnt=i;}
         }
       //  printf("%d\n",ans);
         if(ans==) printf("%d\n",cnt);
         else printf("Not sure\n");
     }
     return ;
 }