鏈接：https://ac.nowcoder.com/acm/contest/887/J
來源：牛客網

題目描述

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: .

輸入描述:

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)

輸出描述:

For each test case, output a single line indicates the reversed sum. 示例1

輸入

復制 3 12 1 101 9 991 1

輸出

復制22 11 2

思路：　　本來想用高精度來求，反噬卻總是wa，只能看大佬代碼 #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; }

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轉載于:https://www.cnblogs.com/RootVount/p/11344588.html

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