hdu 6205 card card card

题目链接

简单题,sub[]记录差值,遍历一遍,用sum记录差值和,小于零的位置做个标记,sum清零,否则继续加和

#include<cstdio>
typedef long long ll;
const int N=1e6+5;
int a[N],b[N],sub[2*N];

int main(){
    int n;
    while(~scanf("%d",&n)){
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
			
		for(int i=1;i<=n;i++){
			scanf("%d",&b[i]);
			sub[i]=a[i]-b[i];
		}
			for(int i=n+1;i<=2*n;i++) sub[i]=sub[i-n]; 
		ll sum=0;
		int res=0;
		for(int i=1;i<=2*n;i++){
			if(sum+sub[i]<0){
				res=i;
				sum=0;
			}
			else
				sum+=sub[i];
		}
		printf("%d\n",res);
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/lidengdengter/article/details/81077437