hdu 5095 Linearization of the kernel functions in SVM (简单模拟)

Linearization of the kernel functions in SVM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2227    Accepted Submission(s): 595

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 + cy^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 + cy^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 + cy^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

20 46 3 4 -5 -22 -8 -32 24 272 31 -5 0 0 12 0 0 -49 12

 

Sample Output

46q+3r+4u-5v-22w-8x-32y+24z+272p+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）

 

虽说是简单模拟，但是有点小坑

中间的规则：

                    1、第一个字母前面的系数如果是正数则不需要输出‘+’；
    2、如果系数是1或-1，那么对于前九项的系数不输出1只有‘+’或‘-’，常数则正常输出；
                    3、如果全部为0时应该输出0，不应该什么也不输出；
                    4、如果前九项的系数为0，最后常数的系数为一个正数，则不需要输出‘+’；*/

#include<stdio.h>#include<string.h>char ss[9]={'p','q','r','u','v','w','x','y','z'};int main(){int n;int map[15];scanf("%d",&n);while(n--){for(int i=0;i<10;i++){scanf("%d",&map[i]);}int count=0;    for(int i=0;i<9;i++)    {     if(map[i]==0)     {     count++;     continue;     }     if(map[i]>0)     {     if(count==i) //前边都是零 {   if(map[i]==1)       {       printf("%c",ss[i]);       }        else       {          printf("%d%c",map[i],ss[i]);          }     }      else     {      if(map[i]==1)          {          printf("+%c",ss[i]);          }           else          {             printf("+%d%c",map[i],ss[i]);             }     }     }     else if(map[i]<0)     {     if(count==i) //前边都是零 {   if(map[i]==-1)       {       printf("-%c",ss[i]);       }        else       {          printf("%d%c",map[i],ss[i]);          }     }      else     {      if(map[i]==-1)          {          printf("-%c",ss[i]);          }           else          {             printf("%d%c",map[i],ss[i]);             }     }     }              }    if(count==9)    {    if(map[9]==0)    {    printf("0\n");    continue;    }    else    {    printf("%d\n",map[9]);    continue;    }    }    else    {     if(map[9]==0)     {     printf("\n");     continue;     }     else     {     if(map[9]>0)       printf("+%d\n",map[9]);     else       printf("%d\n",map[9]);continue;     }    }    }return 0;}