$$(a + b) = a + b$$ $$(a + b)^2\\ =(a+b)a+(a+b)b\\ =a^2+ab+ab+b^2\\ =a^2+2ab+b^2 $$$$(a+b)^3\\ =(a+b)\cdot(a+b)\cdot(a+b)\\ =(a^2+2ab+b^2)\cdot(a+b)\\ =(a^2\cdot a+2ab\cdot a+b^2\cdot a)+(a^2\cdot b+2ab\cdot b+b^2\cdot b)\\ =a^3+3a^2b+3ab^2+b^3 $$$$(a+b)^4\\ =(a^3+3a^2b+3ab^2+b^3)(a+b)\\ =a^4+4a^3b+6a^2b^2+4ab^3+b^4 $$

5 的话同上，结果样例都有了，不解释了。

观察一下系数，珂以发现这是杨辉三角

然后就好了

如果想不到珂以去 oeis

注意数据很坑，需要开 unsigned long long

#include "bits/stdc++.h"
using namespace std;

unsigned long long f[1919][810];

int main() {
    int n;
    scanf ("%d", &n);
    f[1][1] = 1ll;
    f[1][2] = 1ll;
    for (int i = 2; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            f[i][j] = f[i - 1][j - 1] + f[i - 1][j];
        }
    }
    if (n == 1) {printf ("(a+b)^1=a+b"); return 0;}
    printf ("(a+b)^%d=a^%d", n, n);
    for (int i = 1; i < n; ++i) {
        if (n - i == 1) printf ("+%llua", f[n][i + 1]);
        else printf ("+%llua^%d", f[n][i + 1], n - i);
        if (i == 1) printf ("b");
        else printf ("b^%d", i);
    }
    printf ("+b^%d", n);
}