Halloween treats

和POJ2356差不多。

其实这样的数列可以有很多，也可以有不连续的，不过利用鸽巢原理就是方便找到了连续的数列，而且有这样的数列也必定可以找到。

#include 
  
   
#include 
   
     #include 
    
      int main() { int c, n; while (scanf("%d %d", &c, &n) && c) { int *neighbours = (int *) malloc(sizeof(int) * n); int *sumMod = (int *) malloc(sizeof(int) * (n+1)); int *iiMap = (int *) malloc(sizeof(int) * c); std::fill(iiMap, iiMap+c, -1); sumMod[0] = 0; int L = -1, R = -1; for (int i = 0; i < n; i++) scanf("%d", &neighbours[i]); for (int i = 0; i < n; i++) { sumMod[i+1] = (sumMod[i] + neighbours[i]) % c; if (sumMod[i+1] == 0) { L = 1, R = ++i;//下标从1起 break; } if (iiMap[sumMod[i+1]] != -1) { L = iiMap[sumMod[i+1]] + 2, R = ++i; //下标从1起 break; } iiMap[sumMod[i+1]] = i; } if (R != -1) { for (int i = L; i < R; i++) { printf("%d ", i); } printf("%d\n", R); } else puts("no sweets"); free(neighbours), free(iiMap), free(sumMod); } return 0; }