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 … DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN 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:
|
1 2 3 4 5 |
5 1 2 4 14 9 3 1 3 2 5 4 1 |
Sample Output:
|
1 2 3 |
3 10 7 |
数组连续片段和
代码如下：
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 |
#include<stdio.h> #include<stdlib.h> #define ABS(X) ((X)>0?(X):-(X)) int main() { int n,i,m,*distance,*record,total=0,from,to,temp; scanf("%d",&n); distance=(int *)malloc(n*sizeof(int)); record=(int *)malloc(n*sizeof(int)); for(i=0;i<n;i++) { scanf("%d",distance+i); record[i]=total; total+=distance[i]; } scanf("%d",&m); for(i=0;i<m;i++) { scanf("%d%d",&from,&to); temp=ABS(record[from-1]-record[to-1]); if(temp>total-temp) { printf("%d\n",total-temp); } else { printf("%d\n",temp); } } } |