1065. A+B and C (64bit) (20) 大数据溢出问题

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:

31 2 32 3 49223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: falseCase #2: trueCase #3: false

题意：

给定a，b，c三个数字，要求判断a+b的值和c相比是大还是小，如果是大，输出true，否则输出false

分析：

题目给定的数的范围会出现越界，a，b，c都要使用long long型,sum = a+b

当判断越界的时候：

1.当a>0,b>0,sum<0，此时就是sum越界，但是c始终在long long型的范围中，所以此时sum肯定是大于c的，输出true

2.当a<0,b<0,sum>0,此时也是sum越界，说明sum足够小，使得sum越过最小界，但是c始终在范围之中，所以c大于sum，输出false

3.除了这两种情况，剩下来的就是不越界的情况，只需要判断sum和c的大小就可以

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <set>#include <map>using namespace std;const int maxn = 1e5+10;int main(){    int n;    cin >> n;    long long sum,a,b,c;    int cnt = 1;    while(n--)    {       cin >> a >> b >> c;       sum = a+b;       if(a > 0&&b > 0&&sum < 0)       printf("Case #%d: true\n",cnt++);       else if(a < 0&&b < 0&&sum >= 0)       printf("Case #%d: false\n",cnt++);       else if(sum > c)       printf("Case #%d: true\n",cnt++);       else        printf("Case #%d: false\n",cnt++);    }return 0;}