C. Mike and Chocolate Thieves time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! As…

题目大意:有四个小偷,第一个小偷偷a个巧克力,后面几个小偷依次偷a*k,a*k*k,a*k*k*k个巧克力,现在知道小偷有n中偷法,求在这n种偷法中偷得最多的小偷的所偷的最小值. 题目思路:二分查找偷得最多的小偷所偷的数目,并遍历k获取该数目下的方案数.脑子一抽将最右端初始化做了1e15,wa了n多次-- #include<iostream> #include<algorithm> #include<cstring> #include<vector> #in…

Mike and Chocolate Thieves 题目链接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/G Description Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! Aside from loving sweet things, thieves fr…

题目链接: C. Mike and Chocolate Thieves time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrib…

C. Mike and Chocolate Thieves 题目连接: http://www.codeforces.com/contest/689/problem/C Description Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! Aside from loving sweet things, thieves from t…

C. Mike and Chocolate Thieves time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! As…

原题: Description Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! Aside from loving sweet things, thieves from this area are known to be very greedy. So after a thief takes his number of choco…

题目并不难,就是比赛的时候没敢去二分,也算是一个告诫,应该敢于思考…… #include<stdio.h> #include<iostream> using namespace std; int main() { long long n; scanf("%I64d",&n); ,right=1e18,mid,num,m,s; ; while(left<=right) { mid=(left+right)>>; num = ; ;i &l…

题目链接:http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=412145 题目大意:给定一个数字n,问能不能求得一个最小的整数x,使得在1-x范围内,可以取的n组T*k^3<=x.T,k为任意整数,需要在小于等于x的情况下,凑够n对T,k,使其满足T*k^3<=x.若x存在,则输出x,若x不存在,则输出-1. 解题思路: 先看k=2,(1,2,4,8)/(2,4,8,16)/(3,6,12,24).... k=3,(1…

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u   Description Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! Aside from loving sweet things, thieves from this…

因为没注意到long long 就 TLE 了... 二分一下答案就Ok了.. ------------------------------------------------------------------------------ #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0…

题目链接 Solution 先直接二分答案,然后贪心判断,一旦少于答案就吃一块. 思路很简单,有一点细节. 一天内可以不吃巧克力. 注意处理最后时没吃完的全部在最后一天吃完. Code #include<bits/stdc++.h> #define ll long long #define N 50008 #define inf 0x3f3f3f3f3f3f3f using namespace std; void in(ll &x) { char ch=getchar();ll f=1…

C - Mike and Chocolate Thieves Description Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible! Aside from loving sweet things, thieves from this area are known to be very greedy. So after a thie…

A题((Mike and Cellphone) 看起来好像需要模拟数字键位的运动,可是,只要判断出那些必然YES的数字组合不就好了么 #include <cstdio> #include <iostream> #include <cstring> using namespace std; ]; inline bool judge() { ]&&vis[]) return true; ]&&vis[]) return true; ]&…

A - Mike and Cellphone 问有没有多解,每个点按照给出的序列用向量法跑一遍 #include<cstdio> #include<cstring> #include<queue> #include<cstdlib> #include<algorithm> #include<vector> #include<cmath> #include<map> using namespace std; ty…

A 脑筋急转弯 // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <cstdio> #include <cstring> #include <sstream> #include <string> #include <algorithm> #include <list> #incl…

大神们都在刷usaco,我也来水一水 1606: [Usaco2008 Dec]Hay For Sale 购买干草   裸背包 1607: [Usaco2008 Dec]Patting Heads 轻拍牛头 神转化,筛法 1609: [Usaco2008 Feb]Eating Together麻烦的聚餐  LIS 1610: [Usaco2008 Feb]Line连线游戏 排序 1611: [Usaco2008 Feb]Meteor Shower流星雨  BFS 1612: [Usaco2008…

bzoj上的usaco题目还是很好的(我被虐的很惨. 有必要总结整理一下. 1592: [Usaco2008 Feb]Making the Grade 路面修整 一开始没有想到离散化.然后离散化之后就很好做了.F[I,j]表示第i个点,高度>=j或<=j,f[I,j]=min(f[i-1,j]+abs(b[j]-a[i]),f[I,j-1]) 1593: [Usaco2008 Feb]Hotel 旅馆 线段树 ★1594: [Usaco2008 Jan]猜数游戏 二分答案然后写线段树维护 15…

http://www.lydsy.com/JudgeOnline/problem.php?id=2016 这些最大最小显然是二分. 但是二分细节挺多的...这里注意二分的区间,可以累计所有的可能,然后这就是二分区间的右界..(我是sb) 然后二分的时候,判断那里一定要仔细啊.. 还有这题要开longlong啊(雾) #include <cstdio> #include <cstring> #include <cmath> #include <string>…

二分答案,贪心判断,洛谷上要开long long #include<iostream> #include<cstdio> using namespace std; const int N=50005; int n,m,a[N],b[N]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getchar(); } while(p>='0'&…

题目链接:https://ac.nowcoder.com/acm/contest/1577/K 题目大意: 给出n块巧克力,m天吃完.每块巧克力有a[i]快乐值,每天可以选择吃任意块或者不吃巧克力(按照给出的巧克力顺序吃),求如何吃巧克力可以使快乐值的最小值最大. 题解思路: 1.二分最小的开心值.check检查是否有方案可以使得每天的快乐值大于当前的mid,若存在,则答案可以增大,即l = mid + 1.若不存在,则答案减小,即r = mid - 1. 2.对于最后得出来的答案,再进行一次g…

转自: http://blog.csdn.net/qq_26071477/article/details/51892995 #include<stdio.h> typedef long long ll; ll check(ll n) { ll res=0; for(ll k=2; k*k*k<=n; k++)//k是ll范围 不可犯糊涂 res+=n/(k*k*k); return res; } int main() { ll m,res=-1,l=1,r=1e18,mid; scanf…

Friends and Subsequences 题目链接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/H Description Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on the…

C. Read Time Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343/problem/C Description Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data i…

题目 2016: [Usaco2010]Chocolate Eating Time Limit: 10 Sec  Memory Limit: 162 MB Description 贝西从大牛那里收到了N块巧克力.她不想把它们马上吃完,而是打算制定一个计划, 使得在接下来的D天里,她能够尽量地快乐.贝西的快乐指数可以用一个整数来衡量,一开始的时候是0,当她每天晚上睡觉的时候,快乐指数会减半(奇数时向下取整).贝西把她的巧克力按照收到的时间排序,并坚持按照这个顺序来吃巧克力.当她吃掉第i块巧克力的时…

2016: [Usaco2010]Chocolate Eating Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 224  Solved: 87[Submit][Status][Discuss] Description 贝西从大牛那里收到了N块巧克力.她不想把它们马上吃完,而是打算制定一个计划, 使得在接下来的D天里,她能够尽量地快乐.贝西的快乐指数可以用一个整数来衡量,一开始的时候是0,当她每天晚上睡觉的时候,快乐指数会减半(奇数时向下取整).…

D. Friends and Subsequences 题目连接: http://www.codeforces.com/contest/689/problem/D Description Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their o…

Break the Chocolate Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4112 Description Benjamin is going to host a party for his big promotion coming up.Every party needs candies, chocolates and beer, and of cour…

P2985 [USACO10FEB]吃巧克力Chocolate Eating 题目描述 Bessie has received N (1 <= N <= 50,000) chocolates from the bulls, but doesn't want to eat them too quickly, so she wants to plan out her chocolate eating schedule for the next D (1 <= D <= 50,000)…

描述 Bessie has baked a rectangular brownie that can be thought of as an RxC grid (1 <= R <= 500; 1 <= C <= 500) of little brownie squares.The square at row i, column j contains N_ij (0 <= N_ij <= 4,000) chocolate chips.Bessie wants to par…