uva_340 - Master-Mind Hints

/**本题的意思是，对应相等的为强，不对应相等的为弱， * 我的做法是每次比较时复制一个目标串，把他和输入的 *串对比，统计对应相等的，然后把值赋为-1，然后再统计不 *对应相等的 */#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int main(int argc, char const *argv[]){int cas, strong, week, *arr, *cmp, t, *arr1, n(0);while(scanf("%d",&cas),cas){printf("Game %d:\n", ++n);arr = new int[cas+1]; cmp = new int[cas+1]; arr1 = new int[cas+1];for(int i=0; i<cas; i++){scanf("%d",&arr[i]);}while(true){t = 0, strong = 0, week = 0;for(int i=0; i<cas; i++){scanf("%d",&cmp[i]);arr1[i] = arr[i];}if(cmp[0]==0) break;for(int i=0; i<cas; i++){if(arr1[i]==cmp[i]){strong++;cmp[i]=-1; arr1[i] = -1;}}for(int i=0; i<cas; i++){for(int j=0; j<cas; j++){if(arr1[i]==cmp[j] && arr1[i]!=-1){week++; cmp[j] = -1;break;}}}printf("    (%d,%d)\n",strong, week);}}return 0;}