PAT甲级练习1044. Shopping in Mars (25)
来源:互联网 发布:免费翻墙安卓软件 编辑:程序博客网 时间:2024/06/05 17:18
1044. Shopping in Mars (25)
Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:
1. Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).
2. Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
3. Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).
Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.
If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=105), the total number of diamonds on the chain, and M (<=108), the amount that the customer has to pay. Then the next line contains N positive numbers D1 ... DN (Di<=103 for all i=1, ..., N) which are the values of the diamonds. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print "i-j" in a line for each pair of i <= j such that Di + ... + Dj = M. Note that if there are more than one solution, all the solutions must be printed in increasing order of i.
If there is no solution, output "i-j" for pairs of i <= j such that Di + ... + Dj > M with (Di + ... + Dj - M) minimized. Again all the solutions must be printed in increasing order of i.
It is guaranteed that the total value of diamonds is sufficient to pay the given amount.
Sample Input 1:16 153 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13Sample Output 1:
1-54-67-811-11Sample Input 2:
5 132 4 5 7 9Sample Output 2:
2-44-5
一开始用了累加sum和vector来保存输出对,但是两个测试点超时了。。。可以建立sum数组,在输入数据时就加好,不必之后每次再去累加。。。后续程序参考http://blog.csdn.net/jtjy568805874/article/details/50857444
#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <map>#include <set>#include <string>#include <string.h>using namespace std;const int MAX=1e5+10;int d[MAX], sum[MAX], dif;bool f[MAX];vector<int> pe, pg;int main(){int n,m ;scanf("%d %d", &n, &m);for(int i=1; i<=n; i++){scanf("%d", &d[i]);sum[i] = sum[i-1] + d[i];}dif = sum[n];int x=1, y=1, tmp;for(x=1, y=1; x<=n; x++){while(sum[y]-sum[x-1]<m && y<=n) y++;if(sum[y]-sum[x-1]>=m) dif = min(sum[y]-sum[x-1], dif);}for(x=1, y=1; x<=n; x++){while(sum[y]-sum[x-1]<dif && y<=n) y++;if(sum[y]-sum[x-1]==dif) printf("%d-%d\n", x, y);}cin>>n;return 0;}
- PAT甲级练习1044. Shopping in Mars (25)
- 【PAT甲级】1044. Shopping in Mars (25)
- PAT甲级.1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25) PAT 甲级
- PAT甲级1044. Shopping in Mars (25)
- PAT 甲级 1044. Shopping in Mars (25)
- PAT甲级 1044. Shopping in Mars (25)
- 1044. Shopping in Mars (25)-PAT甲级真题(二分查找)
- 1044. Shopping in Mars (25)-PAT
- PAT A 1044. Shopping in Mars (25)
- PAT 1044. Shopping in Mars (25)
- PAT 1044. Shopping in Mars (25)
- PAT 1044. Shopping in Mars (25)
- PAT 1044. Shopping in Mars (25)
- 【PAT】1044. Shopping in Mars (25)
- [pat]1044. Shopping in Mars (25)
- PAT 1044. Shopping in Mars (25)
- PAT(A) - 1044. Shopping in Mars (25)
- webpack入门与解析(一)
- lintcode133最长单词(字符串处理easy)
- Ubuntu 使用cron 设置定时启动任务
- Java Web程序结构
- 局部最小值位置练习
- PAT甲级练习1044. Shopping in Mars (25)
- 软件运行与内存的关系
- 终于理解list_entry和list_for_each_entry
- apache2.4虚拟机配置
- leetcode 151. Reverse Words in a String --------- java
- 【JAVA之多线程】2.运行状态+基本方法
- Cocos2d-x Sprite笔记
- ORACLE的ProC用法讲解
- poj1287最小生成树水