不断向下取直到,忽略的数累计到一个百分比,给当前百分比加1. 这道题要避免处理浮点数,用余数来处理,不然会wa 9 #include <iostream> #include <cmath> using namespace std; int g[10001]; int x, n, sum, le; int main() { cin >> n; for (int i = 1; i <= n; i++) { cin >> g[i]; sum += g[i];…

http://acm.sgu.ru/problemset.php?contest=0&volume=1 101 Domino 欧拉路 102 Coprime 枚举/数学方法 103 Traffic Lights 最短路 104 Little Shop of Flowers 动态规划 105 Div 3 找规律 106 The Equation 扩展欧几里德 107 987654321 Problem 找规律 108 Self-numbers II 枚举+筛法递推 109 Magic of Dav…

SGU 297 题意:就是求余数 收获:无 #include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #defi…

538. Emoticons 题目连接: http://acm.sgu.ru/problem.php?contest=0&problem=538 Description A berland national nanochat Bertalk should always stay up-to-date. That's why emoticons highlighting was decided to be introduced. As making emoticons to be highligh…

HDOJ5054 Alice and Bob Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 302    Accepted Submission(s): 229 Problem Description Bob and Alice got separated in the Square, they agreed that if they…

http://www.nnzhp.cn/archives/152 1.max,min,round print(max([3,4.563,3,6,2.5])) #取最大值,可循环参数即可,int类型的,字符串类型不行 print(min(9,4,7,1,0)) #取最小值 print(round(3.43535345,2)) #取即为小数,四舍五入 print(round(1.12345,2)) 2.sorted s = ' res = sorted(s) #对字符串排序,默认升序 print(l…

链接: http://poj.org/problem?id=1584 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/D A Round Peg in a Ground Hole Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4475   Accepted: 1374 Description The DIY Furniture com…

链接: http://poj.org/problem?id=1584 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/D A Round Peg in a Ground Hole Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4475   Accepted: 1374 Description The DIY Furniture com…

在查询分析器中执行:select rand(),可以看到结果会是类似于这样的随机小数:0.36361513486289558,像这样的小数在实际应用中用得不多,一般要取随机数都会取随机整数.那就看下面的两种随机取整数的方法:1.A:select floor(rand()*N) ---生成的数是这样的:12.0 B:select cast( floor(rand()*N) as int) ---生成的数是这样的:12 2.A:select ceiling(rand() * N) ---生成的数是这…

项目中的一个功能模块上用到了标量值函数,函数中又有ceiling()函数的用法,自己找了一些资料,对SQL中这几个函数做一个简单的记录,方便自己学习.有不足之处欢迎拍砖补充 1.round()函数遵循四舍五入原则,用于把数值字段舍入为指定的小数位数 2.floor(value)函数返回小于或等于指定值(value)的最小整数 3.ceiling(value)函数返回大于或等于指定值(value)的最小整数 例如:对于12.9,floor(12.9)返回12:ceiling(12.9)返回13:r…