HDU 5095 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): 206    Accepted Submission(s): 109

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

 

 有 -1 和 1 的情况 ， 考虑清楚就ok了 。

这是一个细心题，一开始写代码的时候还记得要特别注意1和-1的情况。代码写着写着就忘掉了。

一方面可能是平时组队赛，细心度上对队友产生依赖了

另一方面，太相信自己能在写代码的时候还记得这些注意点。以后应该吧脑海里一闪而过的注意点先写在纸上。

#include<iostream>#include<cstdio>using namespace std;char mp[10]={'p','q','r','u','v','w','x','y','z','A'};int main(){int t;cin >> t;while(t--){int flag = -1;int xx;for(int i=0;i<10;i++){scanf("%d",&xx);if(xx!=0){if(i==9){if(flag ==-1)printf("%d",xx);else{if(xx>0)printf("+%d",xx);elseprintf("%d",xx);}}else{if(flag == -1){if(xx!=1&&xx!=-1)printf("%d%c",xx,mp[i]);else{if(xx == 1)printf("%c",mp[i]);elseprintf("-%c",mp[i]);}}else{if(xx>0){if(xx!=1)printf("+%d%c",xx,mp[i]);elseprintf("+%c",mp[i]);}else{if(xx!=-1)printf("%d%c",xx,mp[i]);elseprintf("-%c",mp[i]);}}}flag = 1;}}cout<<endl;}return 0;}