zyq_20030305的博客

#include<stdio.h> int f[500001][21]; int level[500001]; int to[1000001];//边开2倍 int next[1000001]; int head[500001]; int fa[500001]; int sum,tmp; void addedge(int a,int b)//链表存边 { next[++sum]=head[a]; head[a]=sum; to[sum]=b; } void dfs(int x) { for(int

#include<stdio.h> int x[15001],y[15001],z=1e9,idx; int gcd(int x,int y) { while(y) { z=x%y; x=y; y=z; } return x; } int main() { int n,i,j,v; scanf("%d",&n); printf("0/1\n"); for(i=1;i<=n;i++) for(j=1;j<i;j++) { if(gcd(i,j)==1) x

例题1：usaco Big Barn 巨大的牛棚 Big Barn 巨大的牛棚 Description 农夫约翰想要在他的正方形农场上建造一座正方形大牛棚。他讨厌在他的农场中砍树，想找一个能够让他在空旷无树的地方修建牛棚的地方。我们假定，他的农场划分成 N x N 的方格。输入数据中包括有树的方格的列表。你的任务是计算并输出，在他的农场中，不需要砍树却能够修建的最大正方形牛棚

#include<stdio.h> #include<algorithm> using namespace std; int d[201],f[201]; int main() { int a,b,c,i; scanf("%d%d%d",&a,&b,&c); for(i=1;i<=c;i++) scanf("%d",&d[i]); sort(d+1,d+c+1); for(i=1;i<=c;i++) f[i]=d[i]-d[i-1]; int ans=b; ans-=d[1]

#include<stdio.h> #include<string.h> #include<math.h> int a[151][151]; double f[151][151]; double x[151],y[151]; double mia[151]; double distance(int i,int j) { return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])); } double max(double x,double

Description Like everyone else, FJ is always thinking up ways to increase his revenue. To this end, he has set up a series of tolls that the cows will pay when they traverse the cowpaths throughout t...

Description Bessie and her sister Elsie graze in different fields during the day, and in the evening they both want to walk back to the barn to rest. Being clever bovines, they come up with a pla

Description Farmer John has grown so lazy that he no longer wants to continue maintaining the cow paths that currently provide a way to visit each of his N (5 1..N). Each and every pasture is h