一.技术总结 可以开始把每个数都直接相加当前这个位置的存放所有数之前相加的结果,这样就是递增的了,把i,j位置数相减就是他们之间数的和. 需要写一个函数用于查找之间的值,如果有就放返回大于等于这个数的右下标,函数中采用引用的传参方式,就不需要返回值了. 在处理存放下标问题上,如果发现有更小的数时,直接清空之前存放的值. 二.参考代码 #include<algorithm> #include<iostream> #include<vector> using namespa…

1044 Shopping in Mars (25 分)   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 s…

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 o…

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 o…

题目 Shopping in Mars is quite a diferent 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…

题意: 输入一个正整数N和M(N<=1e5,M<=1e8),接下来输入N个正整数(<=1e3),按照升序输出"i-j",i~j的和等于M或者是最小的大于M的数段. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; ]; vector<pair<int,int> >ans; int main()…

1044. Shopping in Mars (25) 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 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…

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 o…

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 o…

分析: 考察二分,简单模拟会超时,优化后时间正好,但二分速度快些,注意以下几点: (1):如果一个序列D1 ... Dn,如果我们计算Di到Dj的和, 那么我们可以计算D1到Dj的和sum1,D1到Di的和sum2, 然后结果就是sum1 - sum2: (2): 那么我们二分则要搜索的就是m + sum[i]的值. #include <iostream> #include <stdio.h> #include <algorithm> #include <cstr…