PAT 1046 Shortest Distance

 

#include<stdio.h>

int main(){
	int n,m,i,j,d[100000];
	scanf("%d", &n);
	for(i=1;i<=n;i++){
		scanf("%d", &d[i]);
	}
	scanf("%d", &m);
	int start,end,temp,sum1,sum2;
	for(j=0;j<m;j++){
		scanf("%d %d", &start, &end);
		sum1 = sum2 = 0;
		temp = start;
		while(temp != end){
			sum1 += d[temp];
			temp++;
			if(temp>n) temp = 1;
		}
		temp = start;
		while(temp != end){
			if(temp==1) temp = n+1;
			sum2 += d[--temp];
		}
		if(sum1 <= sum2){
			printf("%d\n", sum1);
		}else {
			printf("%d\n", sum2);
		}
	}

	return 0;
}

 运行超时。

 

 

#include<stdio.h>

int main(){
	int n,m,i,j,d[100001],total=0,temp;
	scanf("%d", &n);
	//每个点存储的都是D1到该点的距离总和，D1存储环的总和
	for(i=1;i<=n;i++){
		scanf("%d", &temp);
		total += temp;
		d[i+1] = total;
	}
	d[1] = total;
	scanf("%d", &m);
	int start,end,t,sum1,sum2;
	for(j=0;j<m;j++){
		scanf("%d %d", &start, &end);
		if(start > end){
			t = start;
			start = end;
			end = t;
		}
		if(start == 1){
			sum1 = d[end];
			sum2 = d[1] - sum1;
		}else {
			sum1 = d[end] - d[start];
			sum2 = d[1] - sum1;
		}
		if(sum1 <= sum2){
			printf("%d\n", sum1);
		} else {
			printf("%d\n", sum2);
		}
	}

	return 0;
}

  正常。