hdu---5095---Linearization of the kernel functions in SVM

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5095

分析：当a[i]不为0的时候，就有输出;

　输出内容：1.系数为正数且不是第一个位置输出一个'+';

　　　　　　2.当系数为-1时且不是最后一个常数时输出一个'-';

　　　　　　3.系数不是-1或1或者是最后那个常数时输出这个数；

　　　　　　4.最后一个位置没有字符要输出。

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

 

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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#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include<vector>
#include<algorithm>
using namespace std;
#define PI 3.1415926

const int maxn=1000007;
const int INF=0x3f3f3f3f;

char str[10]={"pqruvwxyz"};
int a[10];

int main()
{
    int T;
    scanf("%d", &T);

    while(T--)
    {
        for(int i=0; i<10; i++)
            scanf("%d", &a[i]);

        int f=0;
        for(int i=0; i<10; i++)
        {
            if(a[i])///当a[i]不为0的时候，就有输出;
            {
                if(f && a[i]>0)printf("+");///系数为正数且不是第一个位置输出一个'+';

                if(a[i]==-1 && i!=9)printf("-");///当系数为-1时且不是最后一个常数时输出一个'-';

                if((a[i]!=-1&&a[i]!=1) || (i==9))printf("%d", a[i]);///系数不是-1或1或者是最后那个常数时输出这个数

                if(i!=9)printf("%c", str[i]);///最后一个位置没有字符要输出

                f=1;///标记判断

            }
        }
        printf("
");
    }
    return 0;
}