Gunner II(二分,map,数字转化)】的更多相关文章

Gunner II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1740    Accepted Submission(s): 635 Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting ve…
 Gunner II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1724    Accepted Submission(s): 631 Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting…
题目传送门 /* 题意:查询x的id,每次前排的树倒下 使用lower_bound ()查找高度,f[i]记录第一棵高度为x树的位置,查询后+1(因为有序) */ #include <cstdio> #include <algorithm> #include <cstring> using namespace std; ; const int INF = 0x3f3f3f3f; struct A { int v, id; bool operator < (cons…
Integer to Roman 问题简介:将输入的int类型数字转化为罗马数字 问题详解:罗马数字由七个不同的符号表示:I,V,X,L,C,D和M 符号-数值 I - 1 V - 5 X -10 L - 50 C - 100 D - 500 M - 1000 例如,2用罗马数字写成II,只有两个I加在一起,十二写为XII,解释为X + II, 二十七写成XXVII,即XX + V + II, 罗马数字通常从左到右从最大到最小,但是,四个数字不是IIII,相反,第四个写为IV,因为一个在五个之前…
本篇阅读的代码实现了将输入的数字转化成一个列表,输入数字中的每一位按照从左到右的顺序成为列表中的一项. 本篇阅读的代码片段来自于30-seconds-of-python. digitize def digitize(n): return list(map(int, str(n))) # EXAMPLES digitize(123) # [1, 2, 3] 该函数的主体逻辑是先将输入的数字转化成字符串,再使用map函数将字符串按次序转花成int类型,最后转化成list. 为什么输入的数字经过这种转…
Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,…
You are given two linked lists representing two non-negative numbers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the two numbers do not con…
//数字转化为汉字 如5-->五-(NSString*)translation:(NSString *)arebic{   NSString *str = arebic;    NSArray *arabic_numerals = @[@"1",@"2",@"3",@"4",@"5",@"6",@"7",@"8",@"9"…
//函数fun功能:将a和b所指的两个字符串分别转化成面值相同的整数,并进行相加作为函数值返回,规定只含有9个以下数字字符. //重难点:字符数字转化为整型数字. #include <stdio.h> #include <string.h> #include <ctype.h> #define N 9 long ctod( char *s ) { ; while(*s)//指针指向字符串首地址. if(isdigit( *s)) {//此函数检查参数是否为字符整数类型.…
剑指 Offer 56 - II. 数组中数字出现的次数 II Offer_56_2 题目详情 解题思路 java代码 package com.walegarrett.offer; /** * @Author WaleGarrett * @Date 2021/2/10 13:43 */ /** * 题目描述:在一个数组 nums 中除一个数字只出现一次之外,其他数字都出现了三次.请找出那个只出现一次的数字. */ public class Offer_56_2 { public int sing…