1065 A+B and C (64bit) （20 分）
Given three integers A, B and C in [−2
63
,2
63
], 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
//這個題在乙級題中已經做過了，當時的一種方案直接拿過來把int改成long long int 直接就通過了
//兩個正數相加溢出爲正數，兩個負數相加溢出爲正數
//正負相加不會溢出
#include <cstdio>
#include <cstring>
#include <cmath>
int main()
{
int T = 0;
long long int A, B, C;
scanf("%d", &T);
for (int i = 1; i <= T; i++)
{
scanf("%lld %lld %lld", &A, &B, &C);
long long int D = A + B;
if (D <= 0 && A>0 && B > 0) printf("Case #%d: true\n", i);//判斷越界時的狀態
else if (D >= 0 && A < 0 && B < 0) printf("Case #%d: false\n", i);
else if (D > C) printf("Case #%d: true\n", i);//否則正常比較輸出即可
else printf("Case #%d: false\n", i);
}
return 0;
}