PAT (Advanced Level) Practise 1065 A+B and C (64bit) (20)

1065. A+B and C (64bit) (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

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：判断两数和与另一个数的关系

解题思路：数据刚好是超了long long int的，所以用了unsigned long long，符号用标记来记录

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;typedef unsigned long long ull;const ull inf = 1ull << 63;char s[1009];ull a, b, c;int main(){int t, cas = 0;scanf("%d", &t);while (t--){printf("Case #%d: ", ++cas);int flag1 = 0, flag2 = 0, flag3 = 0, flag = 1;scanf("%s", s); flag1 = s[0] == '-'; sscanf(flag1 ? s + 1 : s, "%llu", &a);scanf("%s", s); flag2 = s[0] == '-'; sscanf(flag2 ? s + 1 : s, "%llu", &b);scanf("%s", s); flag3 = s[0] == '-'; sscanf(flag3 ? s + 1 : s, "%llu", &c);if (a == inf && a == b && flag1 == flag2&&flag1) flag = 0;else if (flag1 == flag2) b += a;else{if (flag1){if (a > b) flag2 = 1,b = a - b;else b -= a;}else if (a >= b) flag2 = 0, b = a - b;else b -= a;}if (flag2 == flag3){if (flag2&&b >= c) flag = 0;if (!flag2&&b <= c) flag = 0;}else if (flag2) flag = 0;if (flag) printf("true\n");else printf("false\n");}return 0;}