bzoj3907: 网格

链接：http://www.lydsy.com/JudgeOnline/problem.php?id=3907

题意：中文题。。。

分析：用卡特兰数的非常规法证明F[n]=C(2*n,n)-C(2*n,n-1)那种方法可以分析出这题的答案为C(n+m,n)-C(n+m,n+1)，详见百度百科，再用高精度处理一下即可。

代码：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<math.h>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=10010;
const int MAX=151;
const int MOD1=100000007;
const int MOD2=100000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1000000007;
const ll INF=10000000010;
typedef unsigned long long ull;
int a[N],b[N],q[N],num[N];
void deal(int n) {
    int i,j,k=0;
    memset(q,0,sizeof(q));
    for (i=2;i<=n;i++) {
        if (!q[i]) { a[++k]=i;b[i]=i; }
        for (j=1;j<=k;j++) {
            if (a[j]*i>n) break ;
            q[a[j]*i]=1;b[a[j]*i]=a[j];
            if (i%a[j]==0) break ;
        }
    }
}
int ans[N];
void mul(int x) {
    for (int i=1;i<=ans[0];i++) ans[i]*=x;
    for (int i=1;i<ans[0];i++) { ans[i+1]+=ans[i]/10;ans[i]%=10; }
    while (ans[ans[0]]>9) {
        ans[ans[0]+1]=ans[ans[0]]/10;
        ans[ans[0]]%=10;ans[0]++;
    }
}
int main()
{
    int i,j,n,m;
    scanf("%d%d", &n, &m);
    deal(n+m);
    for (i=1;i<=n;i++) num[i]--;
    for (i=1;i<=m;i++) num[i]--;
    for (i=1;i<=n+m;i++) num[i]++;
    num[n-m+1]++;num[n+1]--;
    for (i=n+m;i>1;i--)
    if (!q[i]) continue ;
    else {
        num[b[i]]+=num[i];num[i/b[i]]+=num[i];num[i]=0;
    }
    memset(ans,0,sizeof(ans));
    ans[0]=ans[1]=1;
    for (i=2;i<=n+m;i++)
        for (j=1;j<=num[i];j++) mul(i);
    for (i=ans[0];i>0;i--) printf("%d", ans[i]);
    printf("\n");
    return 0;
}