C++

$\color{gold}{— > 蓝桥杯辅导课题解}$

$40分做法：$

思路：

$暴力$

$直接预处理出来前1000行的杨辉三角形，然后暴力枚举，找到x，输出即可$

$code1:$

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 1e3 + 10;

int n = 1000;
int a[N][N];

int main() {
    a[1][1] = 1;
    for (int i = 2; i <= n; i ++) // 预处理
        for (int j = 1; j <= i; j ++)
            a[i][j] = a[i - 1][j] + a[i - 1][j - 1];

    int x; cin >> x;

    int cnt = 0;
    for (int i = 1; i <= n; i ++) // 枚举
        for (int j = 1; j <= i; j ++) {
            cnt ++;
            if (a[i][j] == x) {
                cout << cnt;
                return 0;
            }
        }       
    return 0;
}

$50分做法：$

思路：

$推公式$

$code2:$

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

typedef long long ll; 

int main() {
    ll n;
    cin >> n;
    cout << n * (n + 1) / 2 + 2;
    return 0;
} 

$80分做法：$

思路：

$将上面两种方法相结合$

$code3:$

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 1e3 + 10;
typedef long long ll; 

int n = 1000;
int a[N][N];

int main() {
    ll x; cin >> x;
    if (x > 1000 * 999 / 2) {
        cout << x * (x + 1) / 2 + 2;
        return 0;
    }

    a[1][1] = 1;
    for (int i = 2; i <= n; i ++)
        for (int j = 1; j <= i; j ++)
            a[i][j] = a[i - 1][j] + a[i - 1][j - 1];

    ll cnt = 0;
    for (int i = 1; i <= n; i ++) 
        for (int j = 1; j <= i; j ++) {
            cnt ++;
            if (a[i][j] == x) {
                cout << cnt;
                return 0;
            }
        }       
    return 0;
}

$满分（100分）做法：$

思路：

二分

$code4:$

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

typedef long long ll; 

int n; 

ll C(int a, int b) { // 求组合数函数 
    ll res = 1;
    for (int i = a, j = 1; j <= b; i --, j ++) {
        res = res * i / j;
        if (res > n) return res;
    }
    return res;
} 

bool check(int k) {
    ll l = k * 2, r = max(n, l);
    while (l < r) {
        ll mid = l + r >> 1;
        if (C(mid, k) >= n) r = mid;
        else l = mid + 1;
    }
    if (C(r, k) != n) return false;

    cout << r * (r + 1) / 2 + k + 1;

    return true;
}

int main() {
    cin >> n;
    for (int k = 16; ; k --)
        if (check(k))
            break;      
    return 0;
}

$ps：以上，皆为在蓝桥杯官网题库测试结果$

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