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; }