n,m然后给出n个数让你求所有存在的区间[l,r],使得a[l]~a[r]的和为m并且按l的大小顺序输出对应区间.如果不存在和为m的区间段,则输出a[l]~a[r]-m最小的区间段方案. 如果两层for循环l和r的话,会超时,实际上for循环一遍即可. #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <vector> #…

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…

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…

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…

#include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <cmath> #include <queue> using namespace std; ]; ]={','A','B','C'}; int main() { scanf(],&color[],&color[]); printf("#…

双指针. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<algorithm> using namespace std; +; long long a[maxn]; in…

有n个老鼠,第一行给出n个老鼠的重量,第二行给出他们的顺序.1.每一轮分成若干组,每组m个老鼠,不能整除的多余的作为最后一组.2.每组重量最大的进入下一轮.让你给出每只老鼠最后的排名.很简单,用两个数组模拟一下即可order1存储进入当前一轮老鼠的索引顺序order2存储进入下一轮老鼠的索引顺序 如果当前有groups个组,那么会有groups个老鼠进入下一轮,则没有进入下一轮的排名都为groups+1如果只有一个组,那么最大的那个排名即为1. #include <iostream> #inc…

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()…

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…