1065. A+B and C (64bit) (20)-PAT甲级

题目：

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

这道题需要知道一些知识点，long long 的范围是[-2^63,2^63-1)
正数之和为负数或者负数之和为正数就算是溢出
A+B必须放在long long 中存下才能进行比较，不能直接在if中比较。
题外话，至于2的63次方有多大，我简单算一下，2的平均3.5左右次方进一位，63/3.5=18，所以长度大概为18，对比上面给出的例子，估计最后一个数就是2的63次方hhh。

解答：

#include<cstdio>int main(){    int T;    scanf("%d",&T);    for(int i=1;i<=T;i++){        long long a,b,c;        scanf("%lld%lld%lld",&a,&b,&c);        long long res=a+b;        if(a>0&&b>0&&res<0)        printf("Case #%d: true\n",i);        else if(a<0&&b<0&&res>=0)        printf("Case #%d: false\n",i);        else if(res>c)        printf("Case #%d: true\n",i);        else        printf("Case #%d: false\n",i);    }    return 0;}