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POJ 3370 Halloween treats(抽屉原理)
Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too late. To avoid conflicts, the children have decided they will put all sweets together and then divide them evenly among themselves. From last year‘s experience of Halloween they know how many sweets they get from each neighbour. Since they care more about justice than about the number of sweets they get, they want to select a subset of the neighbours to visit, so that in sharing every child receives the same number of sweets. They will not be satisfied if they have any sweets left which cannot be divided. Your job is to help the children and present a solution. Input The input contains several test cases. The last test case is followed by two zeros. Output For each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of ai sweets). If there is no solution where each child gets at least one sweet print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them. Sample Input 4 5 1 2 3 7 5 3 6 7 11 2 5 13 17 0 0 Sample Output 3 5 2 3 4 抽屉原理:具体不多讲了,以为是DP,话说我wa了10多发。最后改改又对了。 这个讲的很好:点击打开链接 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> typedef __int64 LL; using namespace std; const int maxn=100000+100; int a[maxn],sum; int pos[maxn]; int main() { int n,m; while(~scanf("%d%d",&m,&n)) { if(n+m==0) break; for(int i=1;i<=n;i++) scanf("%d",&a[i]); sum=0; memset(pos,-1,sizeof(pos)); pos[0]=0; for(int i=1;i<=n;i++) { sum=(sum+a[i])%m; if(pos[sum]!=-1) { for(int j=pos[sum]+1;j<=i;j++) { printf("%d",j); if(j!=i) printf(" "); } puts(""); break; } pos[sum]=i; } } return 0; } |