题解 | #数字阶梯求和#

关键在于字符串的加法

#include <iostream>
#include <string>
#include "vector"
using namespace std;
string multiply(char a, int n) {//构造n个a的字符串
    string ans;
    for (int i = 0; i < n; i++) {
        ans = a + ans;
    }
    return ans;
}
string add(string a, string b) {
    //首先将两个字符串对齐
    int commonSize = a.size() > b.size() ? a.size() : b.size();
    if (a.size() > b.size()) {
        for (int i = 0; i < commonSize - b.size(); i++)
            b = '0' + b;
    } else {
        for (int i = 0; i < commonSize - a.size(); i++)
            a = '0' + a;
    }
//从低位开始进行带进位相加，取模，进位，结果存入a中
    int carry = 0;
    for (int i = commonSize - 1; i >= 0; i--) {
        int temp = a[i] - '0' + b[i] - '0' + carry;
        a[i] = temp % 10 + '0';
        carry = temp / 10;
    }
    if (carry != 0) a = '1' + a;
    return a;
}
int main() {
    char a;
    int n;
    while (cin >> a >> n) { // 注意 while 处理多个 case
        string ans = "0";
        for (int i = 1; i <= n; i++) {
            ans = add(ans, multiply( a, i));
        }
        cout << ans << endl;
    }
}
// 64 位输出请用 printf("%lld")