题解 | #Old Bill#

#include<stdio.h>

int main() {
    int n,x,j,y,z,p,q;
    while(scanf("%d",&n) != EOF) {
        scanf("%d %d %d",&x,&y,&z);
        int max = 0,flag = 0;
        for(int i = 9;i >= 1;i--) {
            for(j = 9;j >= 0;j-- ){
                if((i*10000 + x * 1000 + y*100 + z * 10 + j) % n == 0) {
                        flag = 1;
                    //if((i*10000 + x * 1000 + y*100 + z * 10 + j)/n > max) {//求出每一只火鸡的价格
                        max = (i*10000 + x * 1000 + y*100 + z * 10 + j) / n;
                        p = i,q = j;
                        break;
                    //}
                }
            }
            if(flag == 1) break;
        }
        if(flag){
            printf("%d %d %d\n",p ,q ,max);
        }
        else {
            printf("0\n");
        }
    }
    return 0;
}