HDU 5095 Linearization of the kernel functions in SVM（模拟）

主题链接：http://acm.hdu.edu.cn/showproblem.php?

pid=5095

Problem Description

SVM（Support Vector Machine）is an important classification tool, which has a wide range of applications in cluster analysis, community division and so on. SVM The kernel functions used in SVM have many forms. Here we only discuss the function of the form f(x,y,z) = ax^2 + by^2 + cz^2 + dxy + eyz + fzx + gx + hy + iz + j. By introducing new variables p, q, r, u, v, w, the linearization of the function f(x,y,z) is realized by setting the correspondence x^2 <-> p, y^2 <-> q, z^2 <-> r, xy <-> u, yz <-> v, zx <-> w and the function f(x,y,z) = ax^2 + by^2 + cz^2 + dxy + eyz + fzx + gx + hy + iz + j can be written as g(p,q,r,u,v,w,x,y,z) = ap + bq + cr + du + ev + fw + gx + hy + iz + j, which is a linear function with 9 variables.

Now your task is to write a program to change f into g.

 

Input

The input of the first line is an integer T, which is the number of test data (T<120). Then T data follows. For each data, there are 10 integer numbers on one line, which are the coefficients and constant a, b, c, d, e, f, g, h, i, j of the function f(x,y,z) = ax^2 + by^2 + cz^2 + dxy + eyz + fzx + gx + hy + iz + j.

 

Output

For each input function, print its correspondent linear function with 9 variables in conventional way on one line.

 

Sample Input

2
0 46 3 4 -5 -22 -8 -32 24 27
2 31 -5 0 0 12 0 0 -49 12

 

Sample Output

46q+3r+4u-5v-22w-8x-32y+24z+27
2p+31q-5r+12w-49z+12

 

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

 

Recommend

hujie   |   We have carefully selected several similar problems for you:  5098 5097 5096 5094 5093 

PS:

一道比較坑的模拟题。

注意1和-1 的情况。

代码例如以下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
    int M;
    int a[17];
    char b[17] = {'#','p','q','r','u','v','w','x','y','z'};
    scanf("%d",&M);
    getchar();
    while(M--)
    {
        for(int k = 1; k <= 10; k++)
        {
            scanf("%d",&a[k]);
        }
        int cont = 0;
        int flag = 0;
        for(int k = 1; k < 10; k++)
        {
            if(a[k]==0)
                continue;
            cont++;
            if(cont == 1)
            {
                if(a[k] != 1 && a[k] != -1)
                    printf("%d%c",a[k],b[k]);
                else if(a[k] == 1)
                    printf("%c",b[k]);
                else if(a[k] == -1)
                    printf("-%c",b[k]);
                flag = 1;
            }
            else
            {
                if(a[k] > 0)
                    printf("+");
                if(a[k] != 1 && a[k] != -1)
                    printf("%d%c",a[k],b[k]);
                else if(a[k] == 1)
                    printf("%c",b[k]);
                else if(a[k] == -1)
                    printf("-%c",b[k]);
                flag = 1;
            }
        }
        if(a[10])
        {
            if(a[10] > 0 && flag)
                printf("+");
            printf("%d",a[10]);
            flag = 1;
        }
        if(!flag)//没有答案
            printf("0");
        printf("\n");
    }
    return 0;
}
/*
99
0 0 0 0 0 0 0 0 0 -1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
-1 0 0 0 0 0 0 0 0 0
-1 -1 -1 -41 -1 -1 -1 -1 -1 -1
-1 5 -2 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1
0 0 0 0 0 -1 -1 -1 -1 -1
0 0 0 0 0 1 1 1 1 1
1 1 1 1 1 0 0 0 0 0
-1 -1 -1 -1 -1 0 0 0 0 0
1 1 1 1 1 1 1 1 1 0
*/

版权声明：本文博客原创文章。博客，未经同意，不得转载。