1065 A+B and C (64bit) (20 分)(模擬)
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
分析:
因為A、B的大小為[-2^63, 2^63],用long long 儲存A和B的值,以及他們相加的值sum:
如果A > 0, B < 0 或者 A < 0, B > 0,sum是不可能溢位的
如果A > 0, B > 0,sum可能會溢位,sum範圍理應為(0, 2^64 – 2],溢位得到的結果應該是[-2^63, -2]是個負數,所以sum < 0時候說明溢位了
如果A < 0, B < 0,sum可能會溢位,同理,sum溢位後結果是大於0的,所以sum > 0 說明溢位了
原文連結:https://blog.csdn.net/liuchuo/article/details/52109211
題解
注意用用cin讀取資料最後一個樣例報錯~
#include <bits/stdc++.h> using namespace std; int main() { #ifdef ONLINE_JUDGE #else freopen("1.txt", "r", stdin); #endif int n; cin>>n; for(int i=1;i<=n;i++){ long long int a,b,c; scanf("%lld %lld %lld", &a, &b, &c); long long int sum=a+b; if(a>0&&b>0&&sum<0){ printf("Case #%d: true\n",i); }else if(a<0&&b<0&&sum>=0) printf("Case #%d: false\n",i); else if(sum>c) printf("Case #%d: true\n",i); else printf("Case #%d: false\n",i); } return 0; }
本文來自部落格園,作者:勇往直前的力量,轉載請註明原文連結:https://www.cnblogs.com/moonlight1999/p/15613677.html