PAT-PAT (Advanced Level) Practise A+B and C (64bit) (20) 【二星级】

题目链接：http://www.patest.cn/contests/pat-a-practise/1065

题面：

1065. A+B and C (64bit) (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

Given three integers A, B and C in [-263, 263], you are supposed to tell whether A+B > C.

Input Specification:

The first line of the input gives the positive number of test cases, T (<=10). Then T test cases follow, each consists of a single line containing three integers A, B and C, separated by single spaces.

Output Specification:

For each test case, output in one line "Case #X: true" if A+B>C, or "Case #X: false" otherwise, where X is the case number (starting from 1).
Sample Input:

3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: false
Case #2: true
Case #3: false

题目大意：

判断A+B是否大于C。

解题：

因为直接计算会爆long long，先减再加也不行，那么就取除10部分计算，模10部分分别计算。最后把结果整理一下判断即可。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
#define LL long long
int main()
{
int n;
LL a,ab1,ab2,b,c,c1,c2,t1,t2;
cin>>n;
for(int i=1;i<=n;i++)
{
cout<<"Case #"<<i<<": ";
cin>>a>>b>>c;
ab1=a/10+b/10;
ab2=a%10+b%10;
c1=c/10;
c2=c%10;
t1=ab1-c1;
t2=ab2-c2;
t1+=(t2/10);
t2%=10;
if(t2<0)
t2+=10,t1-=1;
if(t1<0||(t1==0&&t2==0))cout<<"false\n";
else cout<<"true\n";
}
return 0;
}