1065 A+B and C (64bit)

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

作者: HOU, Qiming

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

法一：用高精度去计算

#include <bits/stdc++.h>
using namespace std;int main(){int n=0;cin>>n;for(int j=1;j<=n;j++){string a,b,c,d;cin>>a>>b>>c;if(a.size()<b.size())swap(a, b);vector<int>ai(0),bi(0),ci(0);int fa=1,fb=1,fc=1; //1为正数，0为负数if(a[0]=='-')fa=0;if(b[0]=='-')fb=0;if(c[0]=='-')fc=0;for(int i=a.size()-1;i>=0;i--){if(fa)ai.push_back(a[i]-'0');else if(i!=0)ai.push_back(-(a[i]-'0'));}for(int i=b.size()-1;i>=0;i--){if(fb)bi.push_back(b[i]-'0');else if(i!=0)bi.push_back(-(b[i]-'0'));}for(int i=0;i<bi.size();i++)ai[i]+=bi[i];for(int i=0;i<ai.size();i++)if(ai[i]<0&&i!=ai.size()-1){ai[i]+=10;ai[i+1]--;}else if(ai[i]>9&&i<ai.size()-1){ai[i]-=10;ai[i+1]++;}else if(i==ai.size()-1&&ai[i]>9){ai[i]-=10;ai.push_back(1);}if(ai[ai.size()-1]<0)fa=0;if(fa==1&&fc==0){cout<<"Case #"<<j<<": true"<<endl;continue;}else if(fa==0&&fc==1){cout<<"Case #"<<j<<": false"<<endl;continue;}else if(fa==fc&&fa==1){if(ai.size()>c.size()){cout<<"Case #"<<j<<": true"<<endl;continue;}else if(ai.size()<c.size()){cout<<"Case #"<<j<<": false"<<endl;continue;}int i=0;while (ai[ai.size()-1-i]==c[i]-'0'&&i<ai.size()) i++;if(i<c.size()&&ai[ai.size()-1-i]>c[i]-'0'){cout<<"Case #"<<j<<": true"<<endl;continue;}else {cout<<"Case #"<<j<<": false"<<endl;continue;}}else {if(ai.size()>c.size()){cout<<"Case #"<<j<<": false"<<endl;continue;}else if(ai.size()<c.size()){cout<<"Case #"<<j<<": true"<<endl;continue;}int i=0;while (ai[ai.size()-1-i]==c[i]-'0'&&i<ai.size()) i++;if(i<c.size()&&ai[ai.size()-1-i]<c[i]-'0'){cout<<"Case #"<<j<<": true"<<endl;continue;}else {cout<<"Case #"<<j<<": false"<<endl;continue;}}}return 0;
}

法二：用long long加一点技巧

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
using namespace std;int main()
{int n;scanf("%d",&n);for(int i=1;i<=n;i++){long long a,b,c;scanf("%lld %lld %lld",&a,&b,&c);bool istrue;long long res=a+b;if(a>0&&b>0&&res<=0) istrue=true;else if(a<0&&b<0&&res>=0) istrue=false;else if(res>c) istrue=true;else istrue=false;printf("Case #%d: %s\n",i,istrue?"true":"false");}
}

法三：java的BigInteger


import java.math.BigInteger;
import java.util.Scanner;public class Main {public static void main(String[] args) {// TODO Auto-generated method stubScanner sc = new Scanner(System.in);int n = sc.nextInt();for(int i=0;i<n;i++) {BigInteger a = sc.nextBigInteger();BigInteger b = sc.nextBigInteger();BigInteger c = sc.nextBigInteger();a=a.add(b);if(a.compareTo(c)>0)System.out.println("Case #"+(i+1)+": true");else System.out.println("Case #"+(i+1)+": false");}}
}