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…

题目传送门 /* 贪心:排序后,当a[3] > 2 * (a[1] + a[2]), 可以最多的2个,其他的都是1个,ggr,ggb, ggr... ans = a[1] + a[2]; 或先2个+1个,然后k个rgb...r = x + k; g = 2 * (x + z) + k; b = z + k; ans = (x + z) + k = (a[1] + a[2] + a[3]) / 3; 隔了一段时间有做到这题又不会了,看别人的解题报告水平果然没有提升,以后做题要独立思考,看别人的也要…

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…

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…

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(…

二分!!! AC代码例如以下: #include<iostream> #include<cstring> #include<cstdio> #define ll long long using namespace std; ll n,m,k; ll work(ll a) { ll i,j; ll ans=0; for(i=1;i<=n;i++) { j=a/i; if(j>m) j=m; ans+=j; } return ans; } int main()…