//这是yxc dalao的代码 第一次交的 60000过不去
//#define fre yes

#include <cstdio>

const int N = 60000010;

long long ans[N], tt;

void multi(int b) {
    long long t = 0;
    for (int i = 0; i <= tt; i++) {
        ans[i] = ans[i] * b + t;
        t = ans[i] / 1000000000;
        ans[i] %= 1000000000;
    }

    while(t) {
        ans[++tt] = (t % 1000000000);
        t /= 1000000000;
    }
}

void div(int b) {
    long long t = 0;
    for (int i = tt; i >= 0; i--) {
        ans[i] += t * 1000000000;
        t = ans[i] % b;
        ans[i] /= b;
    }

    while(ans[tt] == 0) tt--;
}

int main() {
    static int n;
    scanf("%d", &n);
    tt = 0;
    ans[0] = 1;
    for (int i = n * 2, j = 1; j <= n; i--, j++) {
        multi(i);
        div(j);
    }

    div(n + 1);

    printf("%lld", ans[tt]);
    for (int i = tt - 1; i >= 0; i--) printf("%09lld", ans[i]);
    puts("");
    return 0;
}

然后我加了个inline 和 register 就能过了（hhh
可能这个代码更容易理解吧，如果有小伙伴不知道这两个是什么（可以百度QAQ）

//#define fre yes

#include <cstdio>

const int N = 60000010;

long long ans[N], tt;

inline void multi(int b) {
    long long t = 0;
    for (register int i = 0; i <= tt; i++) {
        ans[i] = ans[i] * b + t;
        t = ans[i] / 1000000000;
        ans[i] %= 1000000000;
    }

    while(t) {
        ans[++tt] = (t % 1000000000);
        t /= 1000000000;
    }
}

inline void div(int b) {
    long long t = 0;
    for (register int i = tt; i >= 0; i--) {
        ans[i] += t * 1000000000;
        t = ans[i] % b;
        ans[i] /= b;
    }

    while(ans[tt] == 0) tt--;
}

int main() {
    static int n;
    scanf("%d", &n);
    tt = 0;
    ans[0] = 1;
    for (register int i = n * 2, j = 1; j <= n; i--, j++) {
        multi(i);
        div(j);
    }

    div(n + 1);

    printf("%lld", ans[tt]);
    for (register int i = tt - 1; i >= 0; i--) printf("%09lld", ans[i]);
    puts("");
    return 0;
}