转载请注明出处:viewmode=contents" target="_blank">http://blog.csdn.net/u012860063?viewmode=contents 题目链接:http://codeforces.com/contest/448/problem/D -----------------------------------------------------------------------------------------------…

 D. Multiplication Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion isn't just charming, he also is very smart. While some of us were learning the multiplication t…

D. Multiplication Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion isn't just charming, he also is very smart. While some of us were learning the multiplication tabl…

主题链接:http://codeforces.com/contest/448/problem/D 思路:用二分法 code: #include<cstdio> #include<cmath> #include<iostream> using namespace std; __int64 n,m,k; __int64 f(__int64 x) { __int64 res=0; for(__int64 i=1;i<=n;i++) { __int64 minn=min(…

A. Multiplication Table Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/577/problem/A Description Let's consider a table consisting of n rows and n columns. The cell located at the intersection of i-th row and j-th column co…

C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion decided to paint his old fence his…

Problem A: A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present - a new glass cup…

A - Rewards 水题,把a累加,然后向上取整(double)a/5,把b累加,然后向上取整(double)b/10,然后判断a+b是不是大于n即可 #include <iostream> #include <vector> #include <algorithm> #include <cmath> using namespace std; int main(){ double a1,a2,a3; double b1,b2,b3; int n; cin…

C题, #include<cstdio> #include<cstring> #include<algorithm> #define maxn 5005 using namespace std; int num[maxn]; int rmq(int l,int r) { <<,tmp=l; for(int i=l;i<=r;i++) { if(ans>num[i]) { ans=num[i]; tmp=i; } } return tmp; } i…

题目链接 题意: n*m的一个乘法表,从小到大排序后,输出第k个数  (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m) 分析: 对于k之前的数,排名小于k:k之后的数大于,那么就能够採用二分. LL n, m, k; LL fun(LL goal) { LL t = 0, ret = 0; while (++t <= m) { ret += min(n, goal / t); } return ret; } LL bin(LL L, LL R, LL goal) { LL M, V…