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…

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()…

https://pintia.cn/problem-sets/994805342720868352/problems/994805439202443264 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…

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…

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> #…

People in Mars represent the colors in their computers in a similar way as the Earth people. That is, a color is represented by a 6-digit number, where the first 2 digits are for Red, the middle 2 digits for Green, and the last 2 digits for Blue. The…

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…

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…

This time you are asked to tell the difference between the lowest grade of all the male students and the highest grade of all the female students. Input Specification: Each input file contains one test case. Each case contains a positive integer N, f…

Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } where 1. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For exampl…

This time, you are supposed to find A×B where A and B are two polynomials. Input Specification: Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N​1​​ a​N​1​​​​ N​2​​ a​N​2​…

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…

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 som…

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 some…

PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (o…

PAT (Advanced Level) Practice 1008 Elevator (20 分) 凌宸1642 题目描述: The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order.…

PAT乙级:1090危险品装箱(25分) 题干 集装箱运输货物时,我们必须特别小心,不能把不相容的货物装在一只箱子里.比如氧化剂绝对不能跟易燃液体同箱,否则很容易造成爆炸. 本题给定一张不相容物品的清单,需要你检查每一张集装箱货品清单,判断它们是否能装在同一只箱子里. 输入格式: 输入第一行给出两个正整数:N (≤104) 是成对的不相容物品的对数:M (≤100) 是集装箱货品清单的单数. 随后数据分两大块给出.第一块有 N 行,每行给出一对不相容的物品.第二块有 M 行,每行给出一箱货物的清…

PAT乙级:1070 结绳 (25分) 题干 给定一段一段的绳子,你需要把它们串成一条绳.每次串连的时候,是把两段绳子对折,再如下图所示套接在一起.这样得到的绳子又被当成是另一段绳子,可以再次对折去跟另一段绳子串连.每次串连后,原来两段绳子的长度就会减半. 给定 N 段绳子的长度,你需要找出它们能串成的绳子的最大长度. 输入格式: 每个输入包含 1 个测试用例.每个测试用例第 1 行给出正整数 N (2≤N≤104):第 2 行给出 N 个正整数,即原始绳段的长度,数字间以空格分隔.所有整数都不…

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…

1040 Longest Symmetric String (25 分)   Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given Is PAT&TAP symmetric?, the longest symmetric sub-string is s PAT&TAP s, hence you must output 11.…

1044 火星数字 (20 分) 火星人是以 13 进制计数的: 地球人的 0 被火星人称为 tret. 地球人数字 1 到 12 的火星文分别为:jan, feb, mar, apr, may, jun, jly, aug, sep, oct, nov, dec. 火星人将进位以后的 12 个高位数字分别称为:tam, hel, maa, huh, tou, kes, hei, elo, syy, lok, mer, jou. 例如地球人的数字 29 翻译成火星文就是 hel mar:而火星文…

1083 List Grades (25 分) Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval. I…

1065 单身狗 (25 分) “单身狗”是中文对于单身人士的一种爱称.本题请你从上万人的大型派对中找出落单的客人,以便给予特殊关爱. 输入格式: 输入第一行给出一个正整数 N(≤ 50 000),是已知夫妻/伴侣的对数:随后 N 行,每行给出一对夫妻/伴侣——为方便起见,每人对应一个 ID 号,为 5 位数字(从 00000 到 99999),ID 间以空格分隔:之后给出一个正整数 M(≤ 10 000),为参加派对的总人数:随后一行给出这 M 位客人的 ID,以空格分隔.题目保证无人重婚或脚…

此题太给其他25分的题丢人了,只值15分 注意要求最终结果最长,而且向下取整 #include<stdio.h> #include<algorithm> using namespace std; float arr[10005]; int main(){ int N;scanf("%d",&N); for(int i=0;i<N;i++)//输入数据 scanf("%f",&arr[i]); sort(arr,arr+N…

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…

1130 Infix Expression (25 分)(找规律.中序遍历) 我是先在CSDN上面发表的这篇文章https://blog.csdn.net/weixin_44385565/article/details/89035813 Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the preced…

1149 Dangerous Goods Packaging(25 分) When shipping goods with containers, we have to be careful not to pack some incompatible goods into the same container, or we might get ourselves in serious trouble. For example, oxidizing agent (氧化剂) must not be…