1 条题解

  • 0
    @ 2024-12-24 10:06:07

    C :

    #include<stdio.h>
    const int MOD = 1000000007;
    int n;
    long long f[1010];
    int main() {
        f[0] = 1;
        for(int i = 1; i <= 1000; i++) {
            for(int j = 0; j < i; j++) {
                f[i] += f[j] * f[i - j - 1] % MOD;
                f[i] %= MOD;
            }
        }
        while(scanf("%d", &n) == 1) {
            printf("%lld\n", f[n]);
        }
        return 0;
    }
    
    

    C++ :

    #include<iostream>
    #include<stdio.h>
    #include<string.h>
    #define oo 1000000007
    #define ll long long
    using namespace std;
        ll n,dp[1010],i,j;
    int main()
    {
        
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for (i=1;i<=1002;i++)
            for (j=0;j<i;j++)
                dp[i]=(dp[i]+dp[i-j-1]*dp[j]%oo)%oo;
        while (scanf("%lld",&n)!=EOF)
        {
            printf("%lld\n",dp[n]);
        }
        return 0;
    }
    
    
    • 1

    信息

    ID
    1954
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者