POJ 3370 Halloween Treats

Title Link: http://poj.org/problem?id=3370

Test instructions: Each group gives you a number C and a number of M, enter the number of M, let you output a number of subscript, these numbers together can be divided into C (c less than equals N), pigeon nest principle is also applicable,

Note that sum may overflow int, and read in with scanf to prevent timeouts.

Code:

#include <stdio.h> #include <string.h> #include <stdlib.h> #include <ctype.h> #include < math.h> #include <iostream> #include <string> #include <queue> #include <stack> #include < vector> #include <algorithm> #define N 100000+100using namespace std;int a[n];long long int sum[n];int yu[n];    struct node{bool k; int POS;}    Q[n];int Main () {int N, C;    int I, J;    int left, right;        while (~SCANF ("%d%d", &c, &n)) {if (c==0 && n==0) break;        BOOL Flag1=false;        BOOL Flag2=false;            for (i=0; i<n; i++) {scanf ("%d", &a[i]);            if (i==0) sum[i]=a[i];        else Sum[i]=sum[i-1]+a[i];        } left=0;        memset (q, 0, sizeof (q));            for (i=0; i<n; i++) {yu[i]=sum[i]%c;            if (yu[i]==0) {flag1=true; right=i; break;             } else{if (Q[YU[I]].K) {flag2=true;       Left=q[yu[i]].pos; right=i;                Break                }else{q[yu[i]].k=true; q[yu[i]].pos=i;            }}} if (Flag1) {//printf ("%d\n", right+1);                for (i=0; i<=right; i++) {if (i==0) printf ("%d", i+1);            else printf ("%d", i+1);        } printf ("\ n");            } else if (Flag2) {//printf ("%d\n", right-left);                for (i=left+1; i<=right; i++) {if (i==left+1) printf ("%d", i+1);            else printf ("%d", i+1);        } printf ("\ n"); }} return 0;}

This is a person on the net to write code, relatively short: http://www.cnblogs.com/ACShiryu/archive/2011/08/09/poj3370.html for reference to learn.

POJ 3370 Halloween Treats