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

主题链接：

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

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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 
    
     #include 
     
      #include 
      
       #include 
       
        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;}/*990 0 0 0 0 0 0 0 0 -10 0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 01 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 01 1 1 1 1 1 1 1 1 1-1 -1 -1 -1 -1 -1 -1 -1 -1 -10 0 0 0 0 -1 -1 -1 -1 -10 0 0 0 0 1 1 1 1 11 1 1 1 1 0 0 0 0 0-1 -1 -1 -1 -1 0 0 0 0 01 1 1 1 1 1 1 1 1 0*/

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