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…

Description Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team. At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), s and t. You n…

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…

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…

B. Suffix Structures Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team. At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), s and…

题意就是将第一个字符串转化为第二个字符串,支持两个操作.一个是删除,一个是更换字符位置. 简单的字符串操作!. AC代码例如以下: #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #define M 50010 #define inf 100000000 using namespace std; char a[1005],b[1005]; int la,lb;…

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 -----------------------------------------------------------------------------------------------…

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…

 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…

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(…

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…

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…

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…

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",&…

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…

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…

E. Divisors 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 integers. Function f(a) returns the following sequence: first all divisors of a1 go in the increasing order, then al…

C. Painting Fence Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no…

[题目链接]:click here~~ [题目大意]:题意:你面前有宽度为1,高度给定的连续木板,每次能够刷一横排或一竖列,问你至少须要刷几次. Sample Input Input 5 2 2 1 2 1 Output 3 Input 2 2 2 Output 2 Input 1 5 Output 1 搜索: // C #ifndef _GLIBCXX_NO_ASSERT #include <cassert> #endif #include <cctype> #include &…

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()…

Codeforces Round #277 (Div. 2) A. Calculating Function time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output For a positive integer n let's define a function f: f(n) =  - 1 + 2 - 3 + .. + ( - 1)n…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…