HDU problem 3877

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本文探讨了一种特殊的加法运算规则，通过实例解释了如何应用此规则进行数值加法，并按照运算结果的降序输出。文章还提供了一个算法实现来帮助理解和解决相关问题。

Special sort

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 331 Accepted Submission(s): 113

Problem Description

We all know a+b=c,but now there is a new rule in equations:
if a>b,then it's equal to [>c];
if a=b,then it's equal to [=c];
if a<b,then it's equal to [<c];
a sample:

1+1=[=2]
2+0=[>2]
0+2=[<2]
1+2=[<3]

and [>c] > [=c] > [<c].
For every [n],
[n+1] > [n] >[n-1].

Input

The input consists of T cases.The first line of the input contains only positive integer T (0<T<=100).Then follows the cases.Each case begins with a line containing exactly one integer N (0<N<=10000).Then there are N lines, each with two integers a and b (0<a,b<2^31-1).

Output

For each case, calculate each value of a+b(as a+b=[?]) and output a+b=[?] in descending order sorted by [?]. If there exists some pair of a,b whose order can not be determined by the rules above, output them with the order of the input. A blank line should be printed after each case.

Sample Input

Sample Output

  
   5+6=[<11]
4+5=[<9]
5+1=[>6]
3+3=[=6]
1+5=[<6]

6+1=[>7]
1+6=[<7]
4+2=[>6]
3+3=[=6]
2+4=[<6]

Source

2011 Multi-University Training Contest 4 - Host by SDU

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#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
#define MAXN 10001
#define in __int64

typedef struct node
{
    in x,y,ans,ii,z;
}no;
no ct[MAXN];
in cmp(no a,no b)
{
    if(a.ans!=b.ans)
        return a.ans>b.ans;
    else if(a.z!=b.z)
        return a.z>b.z;
    else 
        return a.ii<b.ii;
}
int main()
{
    in t;
    scanf("%I64d",&t);
    in n,i;
    while (t--)
    {
        scanf("%I64d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%I64d %I64d",&ct[i].x,&ct[i].y);
            ct[i].ans=ct[i].x+ct[i].y;
            if(ct[i].x>ct[i].y)
                ct[i].z=3;
            if(ct[i].x<ct[i].y)
                ct[i].z=1;
            if(ct[i].x==ct[i].y) 
                ct[i].z=2;
            ct[i].ii=i;
        }
        sort(ct,ct+n,cmp);
        for(i=0;i<n;i++)
        {
               printf("%I64d+%I64d=[",ct[i].x,ct[i].y);
            if(ct[i].z==3)
                  printf(">");
            if(ct[i].z==2)
                    printf("=");
            if(ct[i].z==1)
                  printf("<");
            printf("%I64d]\n",ct[i].ans);
        }
       printf("\n");
    }
    return 0;
}