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()…

http://codeforces.com/contest/478/problem/C C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have r red, g green and b blue balloons. To decorate a single table for…

题目链接: http://codeforces.com/problemset/problem/650/C C. Table Compression time limit per test4 secondsmemory limit per test256 megabytes 问题描述 Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and m…

E. Table Compression 题目连接: http://www.codeforces.com/contest/651/problem/E Description Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now…

C. Table Decorations time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have r red, g green and b blue balloons. To decorate a single table for the banquet you need exactly three balloons.…

题目链接:http://codeforces.com/contest/448/problem/B ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943/ma…

E. Table Compression time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zipalgorithms and many others.…

C. Table Compression Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now developing the new compression algorithm which he wants to name d…

解题报告 四种情况相应以下四组数据. 给两字符串,推断第一个字符串是怎么变到第二个字符串. automaton 去掉随意字符后成功转换 array 改变随意两字符后成功转换 再者是两个都有和两个都没有 #include <iostream> #include <cstdio> #include <cstring> #include <stdlib.h> #include <algorithm> #include <cmath> usi…

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 cupboard with…

CodeForces - 233D 题目大意给你一个n*m 的矩阵,要求你进行涂色,保证每个n*n的矩阵内都有k个点被涂色. 问你一共有多少种涂色方案. n<=100 && m<=1e18 看数据范围感觉是个矩阵快速幂优化的dp,各种想,连状态转移方程都想不出来,我真 鸡儿菜!!!!,这种和概率有关的dp我感觉好蓝啊!!! 思路:显然是个dp,我们另dp[ i ][ j ]表示,到 i 列,一共涂了j个格子的种数, 那么有状态转移方程 dp[ i ][ j ] +=Σ(dp[…

解题报告 意思就是说有n行柜子,放奖杯和奖牌.要求每行柜子要么全是奖杯要么全是奖牌,并且奖杯每行最多5个,奖牌最多10个. 直接把奖杯奖牌各自累加,分别出5和10,向上取整和N比較 #include <iostream> #include <cstdio> #include <cstring> #include <stdlib.h> #include <algorithm> #include <cmath> using namespa…

题目连接:http://codeforces.com/contest/448 A:给你一些奖杯与奖牌让你推断能不能合法的放在给定的架子上.假设能够就是YES否则就是NO. <span style="font-size:18px;">#include <algorithm> #include <iostream> #include <stdlib.h> #include <string.h> #include <ioman…

题目链接:http://codeforces.com/problemset/problem/448/C 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. B…

E. Divisors time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Bizon the Champion isn't just friendly, he also is a rigorous coder. Let's define function f(a), where a is a sequence of intege…

C. Table Decorations   You have r red, g green and b blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number t of tables can b…

Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now developing the new compression algorithm which he wants to name dis. Petya decided to…

#include <iostream> #include <vector> using namespace std; int main(){ int n,m; cin >> n >> m; ][]; bool flag = false; ; i < n ; ++ i){ ; j < m ; ++ j){ cin >> a[i][j]; } } ; j < m; ++ j){ ][j] || a[n-][j]){ flag = t…

对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespace std; ],b[]; int main(void) { int n; ; ,sum2 = ; ;i<=;++i){ scanf("%d",&a[i]); sum1 += a[i]; } ;i<=;++i){ scanf("%d",&b[…

解题报告 给篱笆上色,要求步骤最少,篱笆怎么上色应该懂吧,.,刷子能够在横着和竖着刷,不能跳着刷,,, 假设是竖着刷,应当是篱笆的条数,横着刷的话.就是刷完最短木板的长度,再接着考虑没有刷的木板,,. 递归调用,,. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define inf 999999999999999 using namespace…

这次CF状态之悲剧,比赛就别提了.后来应该好好总结. A题:某个细节没考虑到,导致T了 代码: #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> using namespace std; int a[4],b[4],n; int main() { while(scanf("%d%d%d",&…