1679: [Usaco2005 Jan]Moo Volume 牛的呼声 Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 723  Solved: 346[Submit][Status] Description Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise.…

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…

Moo Volume Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22104   Accepted: 6692 Description Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise.  FJ's N cows (1 <= N…

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Description Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. FJ's N cows (1 <= N <= 10,000) all…

Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty anim…

Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty anim…

http://www.lydsy.com/JudgeOnline/problem.php?id=1679 水题没啥好说的..自己用笔画画就懂了 将点排序,然后每一次的点到后边点的声音距离和==(n-i)*(a[i+1]-a[i])+之前同样操作所得的的sum 然后答案就是累加后×2 #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iost…

题目链接:http://poj.org/problem?id=2231 思路分析:先排序,再推导计算公式. 代码如下: #include <iostream> #include <algorithm> using namespace std; int main() { , arr[]; int n, m; cin >> n; ; i < n; ++i ) { cin >> arr[i]; } sort( arr, arr + n ); ; i <…

直接枚举两两牛之间的距离即可 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int N=10005; int n,a[N]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getchar(); } while(p>…

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