2019牛客暑期多校训练营（第七场）J A+B problem

2024-05-29 13:24:36

链接：https://ac.nowcoder.com/acm/contest/887/J
来源：牛客网

题目描述

Reversed number is a number written in arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. And all the leading zeros are omitted.
For example: the reversed number of 1234 is 4321. The reversed number of 1000 is 1.

We define reversed number of $x\$ as $f(x)\$ . Given you two positive integers $A\$ and $B\$ , you need to calculate the reversed sum: $f(f(A) + f(B))\$ .

输入描述:

The first line of the input gives the number of test cases, T (T≤300)T\ (T \leq 300)T (T≤300). test cases follow.

For each test case, the only line contains two positive integers: and . (1≤A,B≤231−11 \leq A, B \leq 2^{31}-11≤A,B≤231−1)

输出描述:

For each test case, output a single line indicates the reversed sum.

示例1

输入

复制

输出

复制22

11
2

思路：　　本来想用高精度来求，反噬却总是wa，只能看大佬代码

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;#define ll long long
#define eps 1e-9const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;ll f(ll a)
{ll t = 0;while(a){t = t * 10 + a % 10;a /= 10;}return t;
}ll a, b;int main()
{int t;for(scanf("%d", &t); t--; ){scanf("%lld%lld", &a, &b);printf("%lld\n", f(f(a) + f(b)));}return 0;
}

转载于:https://www.cnblogs.com/RootVount/p/11344588.html

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