1065 A+B and C (64bit) (20 分)大数 溢出
1065 A+B and C (64bit) (20 分)
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
思路:
a、b、c都是long long 范围内的,但是它们相加之后可能会超过范围。
#include<iostream> #include<vector> #include<algorithm> #include<map> #include<set> #include<cmath> #include<climits> using namespace std; int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) { long long a,b,c,x; cin>>a>>b>>c; x=a+b; if(a<0&&b<0&&x>=0) printf("Case #%d: false\n",i+1); else if(a>0&&b>0&&x<0) printf("Case #%d: true\n",i+1); else if(x>c) printf("Case #%d: true\n",i+1); else printf("Case #%d: false\n",i+1); } return 0; }
posted on 2019-01-29 18:30 ZhangのBlog 阅读(602) 评论(0) 编辑 收藏 举报
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