[KTU]Yet another A + B——高精度

Yet another A + B

一、原文

time limit per test : 0.25 s
memory limit per test : 64 MB
input : standard input
output : standard output
You are given three numbers. Is there a way to replace variables A, B and C with these numbers so the equality A + B = C is correct?
Input
There are three numbers X1, X2 and X3 (1 ≤ Xi ≤ 10100), each on a separate line of input.
Output
Output either “YES”, if there is a way to substitute variables A, B and C with given numbers so the equality is correct, or “NO” otherwise.

Examples
input
1 2 3
output
YES
input
1 2 4
output
YES
input
1 3 5
output
NO

二、分析（收获）

1）本题为大数据加法，建议字符数组，利用strcmp好比较。当然结构体和string也是可以的。
2）利用DFS来巧妙处理组合问题。
3）细节上：1、对大数逆置，低位在前利于后续加法和比较
2、通过预处理（比较大小、处理前置0）和*char来传递数组
3、对于最高仍然有进位需要单独处理

三、代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define Max 105
using namespace std;
int flag = 0;
char s[3][Max],ans[3][Max];
char s1[Max],s2[Max];
char *add(char *x,char *y)
{
    if(strlen(x) < strlen(y))
    {
        strcpy(s1,y);
        strcpy(s2,x);
    }
    else
    {
        strcpy(s1,x);
        strcpy(s2,y);
    }
    int d = 0,i = 0;
    for(i = 0;i < strlen(s2);i ++)
    {
        d += s1[i] - '0' + s2[i] - '0';
        s1[i] = d % 10 + '0';
        d /= 10;
    }
    while(s1[i])
    {
        d += s1[i] - '0';
        s1[i ++] = d % 10 + '0';
        d /= 10;
    }
    if(d)
    {
        s1[i ++] = d % 10 + '0';
        d /= 10;
        s1[i] = '\0';
    }
    return s1;
}
void dfs(int k)
{
    if(flag)return ;
    if(k == 3)
    {
        if(strcmp(add(ans[0],ans[1]),ans[2]) == 0)flag = 1;
        return ;
    }
    for(int i = 0;i < 3;i ++)
    {
        strcpy(ans[k],s[i]);
        dfs(k + 1);
    }
}
int main()
{
    for(int i = 0;i < 3;i ++)
    {
        cin>>s[i];
        reverse(s[i],s[i] + strlen(s[i]));
        //由于可能有前置0，需要专门处理；
        for(int j = strlen(s[i]);j >= 1;j --)
        {
            if(s[i][j - 1] != '0')
            {
                s[i][j] = '\0';
                break;
            }
        }
    }
    dfs(0);
    if(flag)cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}