http://pat.zju.edu.cn/contests/pat-a-practise/1046
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:5 1 2 4 14 9 3 1 3 2 5 4 1Sample Output:
3 10 7
拿到题很快就写出了下面的代码,我还在想肿么会这么简单。。。。提交之后超时了,PAT果然到处都是会有坑的= =
#include <cstdio> #define N 100001 int cost[N]; int main(){ int n, m, sum = 0; scanf("%d", &n); for (int i = 1; i <= n; i++){ scanf("%d", &cost[i]); sum += cost[i]; } scanf("%d", &m); while (m--){ int a, b, tmp, len = 0; scanf("%d%d", &a, &b); if (a > b){ tmp = a; a = b; b = tmp; } for (int i = a; i < b; i++){ len += cost[i]; } if (len > sum - len) len = sum - len; printf("%d\n", len); } return 0; }
改进后的代码:
#include <cstdio> #define N 100001 int cost[N]; int main(){ int n, m, sum, cend; scanf("%d", &n); cost[1] = 0; for (int i = 2; i <= n; i++){ int x; scanf("%d", &x); cost[i] = cost[i-1] + x; } scanf("%d", &cend); sum = cost[n] + cend; scanf("%d", &m); while (m--){ int a, b, tmp, len; scanf("%d%d", &a, &b); if (a > b){ tmp = a; a = b; b = tmp; } len = cost[b] - cost[a]; if (len > sum - len) len = sum - len; printf("%d\n", len); } return 0; }