题目链接:传送门 题目大意:就是给四个数,和一个d,问1-d中有多少个数字不是那四个数的倍数; 这道题的d数据很小直接暴力可以过: 暴力代码:时间复杂度O(1): #include<stdio.h> int main(){ int k,m,n,d,l; scanf("%d%d%d%d%d",&k,&l,&m,&n,&d); ; ;i<=d;i++) || i%l== || i%n== || i%m==) ans++; print…

题目链接:http://www.codeforces.com/problemset/problem/148/A题意:求1到d中有多少个数能被k,l,m,n中的至少一个数整出.C++代码: #include <iostream> using namespace std; int k, l, m, n, d, ans; int main() { cin >> k >>l >> m >> n >> d; ; i <= d; i ++)…

公主睡前数龙, 每隔k, l, m, n只都会用不同的技能攻击龙. 假定共数了d只龙, 问共有多少龙被攻击了. 思路: 用一个visit数组记录被攻击过的dragon, 最后遍历visit数组统计被攻击过的龙即可. #define MAXN 100005 using namespace std; int visited[MAXN]; // index from 1 void attack(int c, int d) { int f = d/c; ; i <= f; i++) visited[c*…

Problem description «One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine. However, just counting dragons was boring as well, so she entertained hers…

A. Insomnia cure 哎 只能说英语太差,一眼题我看了三分钟. 题意:给5个数k, l, m, n 和 d,求1~d中能被k, l, m, n 至少一个整除的数的个数. 题解:…… 代码: #include <iostream> using namespace std; int main() { int a, b, c, d, n; cin >> a >> b >> c >> d >> n; ; ; i <= n;…

之所以在codeforces上找这100道水题的原因是为了巩固我对最近学的编程语言的掌握程度. 找的方式在codeforces上的PROBLEMSET中过的题最多的那些题里面出现的最前面的10个题型,它们是: math(数学) brute force(暴力) strings(字符串) implementation(模拟) greedy(贪心) sortings(排序) number theory(数论) constructive algorithm dp(动态规划) shortest path(…

A - Insomnia cure 题意:一共s只龙,每隔k,l,m,n只龙就会受伤,问这s只龙有多少龙是受伤的 思路:看起来题目范围并不是很多,直接进行循环判断 代码: 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7 int main(){ 8…

题目链接:http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1002&cid=723 Cure Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7400    Accepted Submission(s): 1099 Problem Description Given a…

题目链接:hdu5879 Cure 题解:用字符串输入.n很大时答案趋近与(π^2)/6. #include<cstdio> #include<algorithm> #include<iostream> #include<cstring> #include<queue> #include<cmath> using namespace std; ; const double pi = acos(-1.0); char a[N]; dou…

Cure Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1333    Accepted Submission(s): 440 Problem Description Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.   Input…