//就是加减法还没有完成...still CODING#include#include#include#includeconst int MAX= 3000;char *rev(char *a){ //字符串翻转 char re[MAX]; int len = strlen(a); for(int i = 0 ; i <= len-1

又到了JAVA大显神通的时候了...这次用的是BigDecimal类~8行搞定...解决这类问题简直就是手到擒来而如果用C++搞的话,要去前导0,后导0,小数点,再用strcmp()判断是否相等,又复杂又坑爹,在比赛的时候绝对吃亏A == B ?Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/3

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;

Language:DefaultParencodingsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 15029 Accepted: 8959DescriptionLet S = s1 s2...s2n be a well-formed

颤抖吧凡人~这才是真正的暴力枚举!!!枚举16个点的状态分别判断...真的...这样也能过...#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #i

sort横坐标x...以岛屿为圆心.d为半径画圆...分别计算与x轴的左右交点按照从大到小的顺序(或者从小到大都行)对x轴的交点进行判断...#include #include #include #include #include #include #include #include #include #include #include #include #incl

Problem Statement  Everybody who has lived in a village knows how expensive the things from the city are. And everybody who has lived in the city knows how expensive the things from the vill

#include#include#include#include#include#include#include#includeusing namespace std;int main(){ int m,n; cin>>m>>n; if(m==0) if(n==0) cout<<"0"<<" "<<"0"<<endl; else cout<<"Imp

Overflow Write a program that reads an expression consisting of two non-negative integer and an operator. Determine if either integer or the result of the expression is too large to be rep

所有的题目百度都有,问问题前先百度,这是黄金定律所有题都给个思路:具体代码都可百度A:把素数筛出来,先构造原矩形,就是一个一个往下推,然后构造被挖掉合数的矩形遍历原矩形找到起点和终点,然后后构造的矩形中的素数就相当于障碍物,整个问题就化成一个迷宫问题了用BFS一搜就能得出最短距离B:具体证明:http://www.doc88.com/p-792551521271.h