2014上海全国邀请赛1006(hdu 5095)

Linearization of the kernel functions in SVM

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 78 Accepted Submission(s): 63



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

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

题意：直接看样例就能明白题意。
做法：主要要考虑1，-1的情况，0的情况，最前的那个数，如果为正数不用输出+。

#include <iostream>
#include <cstdio>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include<map>
#include <algorithm>
#include<ctime>
#define esp 1e-6
#define LL unsigned long long
#define inf 0x0f0f0f0f
using namespace std;
char fh[20]={"pqruvwxyz"};
int main()
{
int t;
int i,flag,j;
int num[105];
scanf("%d",&t);
while(t--)
{
flag=0;
for(i=0;i<=9;i++)
scanf("%d",&num[i]);
for(i=0;i<=9;i++)
{
if(num[i]!=0)
{
j=i;
break;
}
}
if(j==9||i==10)
{
printf("%d\n",num[9]);
continue;
}
if(num[j]==1)
printf("%c",fh[j]);
else if(num[j]==-1)
printf("-%c",fh[j]);
else
{
if(num[j]>0)
printf("%d%c",num[j],fh[j]);
if(num[j]<0)
printf("%d%c",num[j],fh[j]);
}
for(i=j+1;i<=8;i++)
{
if(num[i]==1)
printf("+%c",fh[i]);
else if(num[i]==-1)
printf("-%c",fh[i]);
else
{
if(num[i]>0)
printf("+%d%c",num[i],fh[i]);
if(num[i]<0)
printf("%d%c",num[i],fh[i]);
}
}
if(num[9]>0)
printf("+");
if(num[9]!=0)
printf("%d",num[9]);
printf("\n");
}
}