PAT(甲级)1046

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3, 10⁵]), followed by N integer distances D₁ D₂ ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_N is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (<=10⁴), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 10⁷.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 931 32 54 1

Sample Output:

3107

#include <cstdio>#include <cstring>const int size = 100005;int d[size];int sum[size];int main(){int N,M,p1,p2;//freopen("test.txt","r",stdin);scanf("%d",&N);//memset(d,0,N*sizeof(int));    scanf("%d",&d[0]);    sum[0] = d[0];for(int i=1;i<N;i++){scanf("%d",&d[i]);sum[i] = sum[i-1]+d[i];}scanf("%d",&M);for(int i = 0;i<M;i++){scanf("%d%d",&p1,&p2);if(p1 > p2){int p =p1;p1= p2;p2 = p;}int d1=0,d2=0;        d1 = sum[p2-1] - sum[p1-1]+d[p1-1] - d[p2-1];if(d1 > sum[N-1] - d1)    d1 =sum[N-1]-d1;printf("%d\n",d1);}//fclose(stdin); return 0;}