hdoj-2095-Linearization of the kernel functions in SVM

Linearization of the kernel functions in SVM

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2239 Accepted Submission(s): 599

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）

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主要是好多细节需要考虑：1. 对于第一个输出的正数的正号是否要输出； 2. 系数为 1或 -1的情况；3 . 0的处理（如果系数都为0，要输出最后一位的0）

#include<stdio.h>char s[10]={'p','q','r','u','v','w','x','y','z'};int main(){int t;scanf("%d",&t);while(t--){int a,k=0;for(int i=0;i<9;++i){scanf("%d",&a);if(a==0) continue;//else k++;if(a==1||a==-1) {if(a==1&&k) printf("+");//else if(a==1&&k==0) ;else if(a==-1)printf("-");printf("%c",s[i]);  k++;continue;}if(a>0&&k) printf("+");printf("%d",a);printf("%c",s[i]);k++;}scanf("%d",&a);                if(a==1||a==-1) {if(a==1&&k==0) printf("1");else if(a==1&&k) printf("+1");else if(a==-1)printf("-1");printf("\n");//continue;}else {if(a==0){if(k==0) printf("0\n");else printf("\n");continue;}if(a>0&&k) printf("+");printf("%d",a);printf("\n");}}return 0;}