题解 | 高精度整数加法

#include <iostream>
using namespace std;

int main() {
    string a, b;
    cin >> a >> b;
    char sum[10002];
    int length = max(a.length(), b.length()) + 1;
    int i, j, Cout;
    for (i = a.length() - 1, j = b.length() - 1, Cout = 0; i >= 0 &&
            j >= 0; --i, --j) {
        if (a[i] + b[j] - 2 * '0' + Cout > 9) {
            sum[--length] = a[i] + b[j] - 2 * '0' + Cout - 10 + '0';
            Cout = 1;
        } else {
            sum[--length] = a[i] + b[j] - 2 * '0' + Cout  + '0';
            Cout = 0;
        }
    }
    while (i >= 0) {
        if (a[i] - '0' + Cout >= 10) {
            sum[--length] = a[i] + Cout - 10;
            Cout = 1;
        } else sum[--length] = a[i] + Cout, Cout = 0;
        i--;
    }
    while (j >= 0) {
        if (b[j] - '0' + Cout >= 10) {
            sum[--length] = b[j] + Cout - 10;
            Cout = 1;
        } else sum[--length] = b[j] + Cout, Cout = 0;
        j--;
    }
    sum[0] = Cout + '0';
    for(i = 0; i < max(a.length(), b.length()) + 1; ++i){
        if(i == 0 && sum[i] != '0'){
            cout<<sum[i];
        }else if(i != 0)
            cout<<sum[i];
    }
}
// 64 位输出请用 printf("%lld")