--这可能是早年Pascal盛行的时候考排序的吧居然还是Glod-- #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int N=100005; int n,a[N]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=ge…

1753: [Usaco2005 qua]Who's in the Middle Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 290  Solved: 242[Submit][Status][Discuss] Description FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives…

Description Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in tw…

原来强行转int可以避免四舍五入啊 #include<iostream> #include<cstdio> using namespace std; int r,y; double m; int main() { scanf("%d%lf%d",&r,&m,&y); double l=1.0+(double)r/100.0; for(int i=1;i<=y;i++) m*=l; printf("%d\n",(i…

高精乘法板子 然而WA了两次也是没救了 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=105; int la,lb,lc,a[N],b[N],c[N],tot; char ch[N]; int main() { scanf("%s",ch+1); la=strlen(ch+1); for(int i=1;i<=la;i…

Description FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cows give as much or more than the median; half give as much or less. Given an odd number of cows N (1 <= N < 10,0…

1755: [Usaco2005 qua]Bank Interest Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 187  Solved: 162[Submit][Status][Discuss] Description Farmer John made a profit last year! He would like to invest it well but wonders how much money he will make. He kn…

1754: [Usaco2005 qua]Bull Math Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 374  Solved: 227[Submit][Status] Description Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... o…

1751: [Usaco2005 qua]Lake Counting Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 168  Solved: 130 [Submit][Status] Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x…

1754: [Usaco2005 qua]Bull Math Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 398  Solved: 242[Submit][Status][Discuss] Description Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answ…

1751: [Usaco2005 qua]Lake Counting Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 190  Solved: 150[Submit][Status][Discuss] Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle…

bzoj3384[Usaco2004 Nov]Apple Catching 接苹果 bzoj1750[Usaco2005 qua]Apple Catching 题意: 两棵树,每分钟会从其中一棵树上掉一个苹果下来,捡苹果的人只愿意W次,问初始在树1处最多能捡多少苹果.分钟数≤1000,W≤30. 题解: dp.f[i][j][0/1]表示第i分钟移动了j次现在在第2/1棵树下.具体看代码. 代码: #include <cstdio> #include <cstring> #incl…

[算法]高精度乘法 #include<cstdio> #include<algorithm> #include<cstring> using namespace std; ; char s1[maxn],s2[maxn]; int a[maxn],b[maxn],c[maxn],lena,lenb,lenc; int main(){ scanf("%s%s",s1,s2); lena=strlen(s1);lenb=strlen(s2); ;i<…

1687: [Usaco2005 Open]Navigating the City 城市交通 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 122  Solved: 85[Submit][Status][Discuss] Description     由于牛奶市场的需求,奶牛必须前往城市,但是唯一可用的交通工具是出租车．教会奶牛如何在城市里打的．     给出一个城市地图,东西街区E(1≤E≤40),南北街区N(1≤N≤30).制作一个开车指南给出…

1674: [Usaco2005]Part Acquisition Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 337  Solved: 162[Submit][Status][Discuss] Description The cows have been sent on a mission through space to acquire a new milking machine for their barn. They are flying…

题目链接:BZOJ - 1733 题目分析 直接二分这个最大边的边权,然后用最大流判断是否可以有 T 的流量. 代码 #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MaxN = 200 + 5,…

题目 1739: [Usaco2005 mar]Space Elevator 太空电梯 Time Limit: 5 Sec  Memory Limit: 64 MB Description The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) differe…

题目 1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚 Time Limit: 5 Sec  Memory Limit: 64 MB Description Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most…

完全背包.. --------------------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; i++ ) #defi…

一开始直接 O( n² ) 暴力..结果就 A 了... USACO 数据是有多弱 = = 先sort , 然后自己再YY一下就能想出来...具体看code ----------------------------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<i…

题目 1684: [Usaco2005 Oct]Close Encounter Time Limit: 5 Sec  Memory Limit: 64 MB Description Lacking even a fifth grade education, the cows are having trouble with a fraction problem from their textbook. Please help them. The problem is simple: Given a…

题目 1677: [Usaco2005 Jan]Sumsets 求和 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 617  Solved: 344[Submit][Status] Description Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers…

一道水题WA了这么多次真是.... 统考终于完 ( 挂 ) 了...可以好好写题了... 先floyd跑出各个点的最短路 , 然后二分答案 m , 再建图. 每个 farm 拆成一个 cow 点和一个 shelter 点, 然后对于每个 farm x : S -> cow( x ) = cow( x ) 数量 , shelter( x ) -> T = shelter( x ) 容量 ; 对于每个dist( u , v ) <= m 的 cow( u ) -> shelter( v…

dp... dp( l , r , k )  , 表示 吃了[ l , r ] 的草 , k = 1 表示最后在 r 处 , k = 0 表示最后在 l 处 . -------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep…

最小最大...又是经典的二分答案做法.. -------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; ++i ) #def…

链接 1735: [Usaco2005 jan]Muddy Fields 泥泞的牧场 思路 这就是个上一篇的稍微麻烦版(是变脸版,其实没麻烦) 用边长为1的模板覆盖地图上的没有长草的土地,不能覆盖草地 每个点(x,y)只有选择x或者y才能被覆盖 还是最小点覆盖,证明在上一篇 横边和竖边得遍历一遍求出,因为不能越过草地嘛 然后左边横边,右边竖边,开心的跑最大流就可以 代码 #include <bits/stdc++.h> #define iter vector<int>::itera…

1674: [Usaco2005]Part Acquisition Time Limit: 5 Sec  Memory Limit: 64 MB Description The cows have been sent on a mission through space to acquire a new milking machine for their barn. They are flying through a cluster of stars containing N (1 <= N <…

1734: [Usaco2005 feb]Aggressive cows 愤怒的牛 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000). His C (2 <= C &l…

1689: [Usaco2005 Open] Muddy roads 泥泞的路 Description Farmer John has a problem: the dirt road from his farm to town has suffered in the recent rainstorms and now contains (1 <= N <= 10,000) mud pools. Farmer John has a collection of wooden planks of…

Description 贝茜想驾驶她的飞船穿过危险的小行星群．小行星群是一个NxN的网格(1≤N≤500),在网格内有K个小行星(1≤K≤10000)． 幸运地是贝茜有一个很强大的武器,一次可以消除所有在一行或一列中的小行星,这种武器很贵,所以她希望尽量地少用．给出所有的小行星的位置,算出贝茜最少需要多少次射击就能消除所有的小行星． Input 第1行:两个整数N和K,用一个空格隔开． 第2行至K+1行:每一行有两个空格隔开的整数R,C(1≤R,C≤N),分别表示小行星所在的行和列． Outpu…