链接:https://ac.nowcoder.com/acm/contest/338/G
来源:牛客网
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld
题目描述
This is a very simple problem! Your only job is to calculate a + b + c + d!
输入描述:
There are several cases. In the first line, there is a single integer T.(T <= 200) In the next T lines, each line contains four integers a, b, c and d(-2^61 <= a,b,c,d <=2^61)
输出描述:
output T lines. Each line output one integer represent the answer of a + b + c + d
示例1
输入
1 1 2 3 4
输出
10
分析:考察对于长整型数据范围的认知,直接加法是会出现错误的,需要在a,b,c,d都等于2^61时特判直接输出2^63,其他情况直接输出结果即可
#include<iostream>
using namespace std;
int main()
{
int c;cin>>n;
while(n--)
{
long long a,b,c,d;
cin>>a>>b>>c>>d;
if(a==2305843009213693952&&b==2305843009213693952&&c==2305843009213693952&&d==2305843009213693952)
cout<<"9223372036854775808"<<endl;
else
cout<<a+b+c+d<<endl;
}
}
当然也可以用Python来解此题(一开始我以为是大数模拟,后来发现用Python就行了)
T=int (raw_input().strip())
for case in range(T):
a,b,c,d=map(int,raw_input().strip().split())
print a+b+c+d