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
as
. Given you two positive integers
and
, you need to calculate the reversed sum:
.
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
3
12 1
101 9
991 1
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-9
const 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
