NC14321. A + B Is Overflow
描述
输入描述
There are multiple test cases. The first line of each test case is a positive integer T, indicating the number of test cases.
For each test case, there is only one line including two 32-bit signed integers A and B.
输出描述
For each test case, output one line. If the sum of A and B will exceed the range of integer, print "Yes", else print "No".
示例1
输入:
3 1 2 -2147483648 2147483647 2147483647 2147483647
输出:
No No Yes
C(clang 3.9) 解法, 执行用时: 1ms, 内存消耗: 296K, 提交时间: 2020-07-26 10:55:52
# include <stdio.h>
int main(){
long long int a,b;
int n,i=0,s;
scanf("%d",&n);
for(;i<n;i++){
scanf("%lld %lld",&a,&b);
s=a+b;
printf("%s\n",(s-a)!=b?"Yes":"No");
}
return 0;
}
C++(g++ 7.5.0) 解法, 执行用时: 3ms, 内存消耗: 384K, 提交时间: 2022-12-01 21:25:55
#include<bits/stdc++.h>
using namespace std;
int main(){
int a;
cin>>a;
long long c,d;
while(a--){
cin>>c>>d;
int sum=c+d;
if(c+d==sum) cout<<"No"<<endl;
else cout<<"Yes"<<endl;
}
}
Python3 解法, 执行用时: 38ms, 内存消耗: 4500K, 提交时间: 2022-10-22 18:25:44
T=int(input())
for _ in range(T):
a,b=map(int,input().split())
if((a+b)>-2147483648 and (a+b)<2147483648):
print('No')
else:print('Yes')