Every year there is the same problem at Halloween: Each neighbour is only willing to give a certaintotal number of sweets on that day, no matter how many children call on him, so it may happen thata child will get nothing if it is too late. To avoid conicts, the children have decided they will putall sweets together and then divide them evenly among themselves. From last year's experience ofHalloween they know how many sweets they get from each neighbour. Since they care more aboutjustice than about the number of sweets they get, they want to select a subset of the neighbours tovisit, so that in sharing every child receives the same number of sweets. They will not be satis ed ifthey have any sweets left which cannot be divided.Your job is to help the children and present a solution.InputThe input contains several test cases.The rst line of each test case contains two integers c and n (1 <= c<= n <= 100000), the number ofchildren and the number of neighbours, respectively. The next line contains n space separated integersa1; : : : ; an (1 <= ai <= 100000), where ai represents the number of sweets the children get if they visitneighbour i.The last test case is followed by two zeros.OutputFor 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 whereeach child gets at least one sweet, print `no sweets' instead. Note that if there are several solutionswhere each child gets at least one sweet, you may print any of them.Sample Input4 51 2 3 7 53 67 11 2 5 13 170 0Sample Output3 52 3 4
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=100000+5;typedef long long LL;LL a[maxn];LL mod[maxn] ;
int main(){ int m,n; while(~scanf("%d%d",&m,&n)) { if(m==0 && n==0) break; memset(a,0,sizeof(a)); int k; for(int i=1; i<=n; i++) //方法一 // scanf("%d",&a[i]);//,mod[i]=-2; //memset(mod, -1, sizeof(mod)); //mod[0]=-1; /*LL sum=0; for(int i=0; i<n; i++) { sum+=a[i]; if(mod[sum%m]!=-2) { for(int j=mod[sum%m]+1; j<=i; j++) { printf("%d",j+1); if(i!=j) printf(" "); } cout<<endl; break; } mod[sum%m]=i; }*/ { scanf("%d",&k); a[i]=a[i-1]+k; } memset(mod, 0, sizeof(mod));
for(int i=1 ; i<=n; i++) { if(a[i]%m == 0) { for(int j=1; j<=i; j++) { if(j == 1) printf("%d",j); else printf(" %d",j); } cout<<endl; break; } if(mod[a[i]%m] == 0) mod[a[i]%m]=i; else { for(int j=mod[a[i]%m]+1; j<=i; j++) { if(j == i) cout<<j<<endl; else cout<<j<<" "; } break; } } } return 0;}
第二讲课件代码模式：
转载于:https://www.cnblogs.com/chen9510/p/4702395.html
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