#4 超时 求调
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#4 超时 求调
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Gary0925楼主2024/7/27 16:43
#include <iostream>
#include <string>
#include<cmath>
using namespace std;

int main()
{
    short int a[510] = { 0 }, b[510] = { 0 };
    int c, d = 0, s;
    string in;
    cin >> in;
    s = in.length();
    for (int i = 0; i < s; i++)
    {
        a[i] = in[s - i - 1] - '0';
    }
    cin >> in;
    for (int i = 0; i < in.length(); i++)
    {
        c = in[in.length() - i - 1] - '0';
        a[i] = (d + c + a[i]);
        d = a[i] / 10;
        a[i] -= 10 * d;
    }
    a[in.length()] = (d + a[in.length()]);
    d = a[in.length()] / 10;
    a[in.length()] -= 10 * d;
    for (int i = 1 + in.length(); i <= s; i++)
    {
        a[i] = (d + a[i]);
        d = a[i] / 10;
        a[i] -= 10 * d;
    }
    for (c = 505; a[c] == 0; c--) {}
    if (c == -1)
    {
        cout << 0;
        return 0;
    }
    for (int i = c; i >= 0; i--)
    {
        cout << a[i];
    }
    return 0;
}
2024/7/27 16:43
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