Kingdom of Obsession

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2799    Accepted Submission(s): 835Problem Description

There is a kindom of obsession, so people in this kingdom do things very strictly. They name themselves in integer, and there are n people with their id continuous (s+1,s+2,⋯,s+n) standing in a line in arbitrary order, be more obsessively, people with id x wants to stand at yth position which satisfy

xmody=0

Is there any way to satisfy everyone's requirement?

Input

First line contains an integer T, which indicates the number of test cases. Every test case contains one line with two integers n, s. Limits 1≤T≤100. 1≤n≤109. 0≤s≤109.

Output

For every test case, you should output 'Case #x: y', where x

Sample Input

2 5 14 4 11

Sample Output

Case #1: No Case #2: Yes

Source

顯然有兩個s+1到s+1+n內有兩個素數肯定輸出No，因為兩個連續的素數的間隔非常小，直接跑二分匹配就行了

/* 題目連結： http://acm.hdu.edu.cn/showproblem.php?pid=5943 
題目大意：　　
給定S，N，把S+1，S+2，...S+N這N個數填到1,2,...,N裡，要求X只能填到X的因子的位置。    
（即X%Y=0，那麼X才能放在Y位置）    問是否能夠放滿。 
題目思路：  【二分圖匹配 匈牙利演算法】 首先，如果S<N,那麼S+1，S+2...N這些數直接放在S+1,S+2...N的位置上
(如果其他數x放在這些位置上面，這些數不放在對應位置，那麼x一定能放在這些數放的位置，所以直接交換即可)
所以可以直接將S和N調換，縮小N。接著看N個連續的數，如果出現2個素數。那麼必然無解(都只能放1)所以可以估算一下素數的最大間隔(我取504),N超過必然無解。
N小於504的情況下，直接暴力建邊（能整除就連邊），然後跑二分圖匹配即可
摘自：http://blog.csdn.net/u010568270/article/details/52965762*/

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1025;
const int MAXM = 1025 * 1025;
struct Edge
{
    int to,next;
}edge[MAXM];
int tot,head[MAXN];
void addedge(int u,int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}
int linker[MAXN];
bool used[MAXN];
int uN;
bool dfs(int u)
{
    for(int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].to;
        if(!used[v]) {
            used[v] = true;
            if(linker[v] == -1 || dfs(linker[v])) {
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}
int hungary()
{
    int res = 0;
    memset(linker,-1,sizeof(linker));
    for(int u = 1; u <= uN; u++) {
        memset(used,false,sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}
int main(void)
{
    int T,kase = 0,n,s;
    scanf("%d",&T);
    while(T--) {
        kase++;
        init();
        scanf("%d %d",&n,&s);
        if(s < n) swap(n,s);
        if(n >= 504) {
            printf("Case #%d: No\n",kase);
            continue;
        }
        else {
            uN = n;
            for(int i = 1; i <= n; i++) {
                for(int j = 1; j <= n; j++) {
                    if((i + s) % j == 0) {
                        addedge(i + n,j);
                        addedge(j,i + n);
                    }
                }
            }
            if(hungary() == n) printf("Case #%d: Yes\n",kase);
            else printf("Case #%d: No\n",kase);
        }
    }
    return 0;
}