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

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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
故意卡在longlong的上限上，可以感受到出题人满满的恶意，先读在字符串里，然后去掉符号用unsignedlonglong来存，剩下的就是判断符号和越界的问题。
#include<cmath>#include<queue>#include<string>#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;typedef unsigned long long ull;const int mod = 1e9 + 7;const int maxn = 1e3 + 10;const ull INF = ((ull)1 << 63);int T, t;char s[maxn];ull a, b, c;bool check(){int f1 = 0, f2 = 0, f3 = 0;scanf("%s", s);f1 = s[0] == '-'; sscanf(f1 ? s + 1 : s, "%llu", &a);scanf("%s", s);f2 = s[0] == '-'; sscanf(f2 ? s + 1 : s, "%llu", &b);scanf("%s", s);f3 = s[0] == '-'; sscanf(f3 ? s + 1 : s, "%llu", &c);if (a == INF && a == b && f1 == f2) return !f1;if (f1 == f2) b += a; else{if (f1) if (a > b) { f2 = 1; b = a - b; } else b -= a;else if (a >= b) { f2 = 0; b = a - b; } else b -= a;}return f2 == f3 ? f2 ? b < c : b > c : f2 < f3;}int main(){scanf("%d", &T);while (T--){printf("Case #%d: %s\n", ++t, check() ? "true" : "false");}return 0;}

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