PAT 1065. A+B and C (64bit) (20) 大整数，有正负，加减。疑问已解决

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

经验教训就是不要妄图用加法搞定一切，写的各种麻烦。不如直接再写出个减法，轻松愉快。

#include<string>#include<stdio.h>#include<iostream>#include<string.h>#include<queue>#include<algorithm>#include<map>#include<set>#include<math.h>#include<stack>#include<vector>using namespace std;struct ZZ{  int num[1000];  int p=0;  int mark;       // 标记正负。1为正，-1为负  void getnum(char x[])   {       p=0;       for(int i=0;i<1000;i++)        num[i]=0;       int l=strlen(x)-1;       int k=1;       int sum=0;       mark=1;       if(x[0]!='-')    {       for(int i=l;i>=0;i--)         {            sum+=(x[i]-'0')*k;            k*=10;             if(k==10000||i==0)             {                 k=1;                 num[p++]=sum;                 sum=0;             }         }    }          else         {            mark=-1;            if(strlen(x)==2&&x[1]=='0') mark=1;           for(int i=l;i>=1;i--)           {             sum+=(x[i]-'0')*k;             k*=10;             if(k==10000||i==1)               {                 k=1;                 num[p++]=sum;                 sum=0;               }            }          }   }    ZZ operator + (const ZZ &A) const    {       ZZ tmp;       tmp.getnum("0");         int pp=0;         int d=0;         while(pp<p||pp<A.p)         {            int sum=num[pp]+A.num[pp]+d;              tmp.num[pp++]=sum%10000;             d=sum/10000;         }           if(d!=0) tmp.num[pp++]=d;            tmp.mark=mark;            tmp.p=pp;                   return tmp;    }      ZZ operator - (const ZZ &A) const       {           ZZ tmp;          tmp.getnum("0");          int pp=0;          int d=0;       while(pp<p)       {           int sum=num[pp]-A.num[pp]-d;           d=0;           if(sum<0) {                         sum+=10000;                          d=1;                     }           tmp.num[pp++]=sum;       }         for(int i=100;i>=0;i--)         if(tmp.num[i]!=0) {pp=i+1;break; }         tmp.p=pp;         tmp.mark=mark;         return tmp;       }     bool postive()    //判断是否为正数      {         int fla=0;         for(int i=0;i<p;i++)           if(num[i]!=0)           {              fla=1;              break;           }          if(fla==1&&mark==1) return true;          return false;      }    bool operator > (const ZZ &A) const   //比较绝对值大小    {        if(p!=A.p) return p>A.p;        for(int i=p-1;i>=0;i--)          if(num[i]!=A.num[i]) return num[i]>A.num[i];           return false;    }};int main(){     char a[70],b[70],c[70];     int n;     cin>>n;     for(int i=1;i<=n;i++)     {       cin>>a>>b>>c;         ZZ t1,t2,t3,t4;         t1.getnum(a);         t2.getnum(b);         t3.getnum(c);         if(t1.mark==t2.mark)         t4=t1+t2;         else         {            if(t1>t2) t4=t1-t2;            else t4=t2-t1;         }         //cout<<t4.mark<<' '<<t4.num[1]<<' '<<t4.num[0]<<' '<<t4.p<<endl;         if(t4.mark!=t3.mark)         t4=t4+t3;         else{               if(t4>t3) t4=t4-t3;                else  {t4=t3-t4;t4.mark=-t3.mark;}             }          //cout<<t4.mark<<' '<<t4.num[1]<<' '<<t4.num[0]<<' '<<t4.p<<endl;         cout<<"Case #"<<i<<": ";         if(t4.postive())            cout<<"true";         else cout<<"false";         cout<<endl;     }     return 0;}

看了人家的方法，羞愧

下面俩仅仅是sum的差别。居然第三个过不了，神奇

评测结果
时间结果得分题目语言用时(ms)内存(kB)用户8月13日 11:26答案正确201065C++ (g++ 4.7.2)1384datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确130812/121答案正确13844/42答案正确13084/4
#include <iostream>  using namespace std; int main(){   int T, index;  long long a, b, c,sum;  bool flag;  cin >> T;  for (index = 0; index < T;)  {    index++;    cin >> a >> b >> c;     sum = a + b;    if (a>0 && b > 0 && sum <= 0)flag = true;    else if(a < 0 && b < 0 && sum>= 0)flag = false;    else if (sum> c)flag = true;    else flag = false;    cout << "Case #" << index << ": " << (flag ? "true" : "false" )<< endl;  }  system("pause");  return 0;}

评测结果
时间结果得分题目语言用时(ms)内存(kB)用户8月13日 11:23部分正确121065C++ (g++ 4.7.2)1436datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确143612/121答案错误13080/42答案错误14360/4
#include <iostream>  using namespace std; int main(){   int T, index;  long long a, b, c;  bool flag;  cin >> T;  for (index = 0; index < T;)  {    index++;    cin >> a >> b >> c;     if (a>0 && b > 0 && (a + b) <= 0)flag = true;    else if(a < 0 && b < 0 && (a + b )>= 0)flag = false;    else if ((a + b )> c)flag = true;    else flag = false;    cout << "Case #" << index << ": " << (flag ? "true" : "false" )<< endl;  }  system("pause");  return 0;}

sum过了而a+b没过是因为，当数据比较大时(即a+b > 2^63-1)此时sum因为规定为long long了，所以会被当成负数。而a+b在2^64和2^63-1 之间会被当做无符号长整型处理。ISO C99里这么规定的