主题链接：http://acm.hdu.edu.cn/showproblem.php?
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 + cz^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 + cz^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.
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 + cz^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
Recommend
PS:
一道比較坑的模拟题。
注意1和-1 的情况。
代码例如以下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
int M;
int a[17];
char b[17] = {'#','p','q','r','u','v','w','x','y','z'};
scanf("%d",&M);
getchar();
while(M--)
{
for(int k = 1; k <= 10; k++)
{
scanf("%d",&a[k]);
}
int cont = 0;
int flag = 0;
for(int k = 1; k < 10; k++)
{
if(a[k]==0)
continue;
cont++;
if(cont == 1)
{
if(a[k] != 1 && a[k] != -1)
printf("%d%c",a[k],b[k]);
else if(a[k] == 1)
printf("%c",b[k]);
else if(a[k] == -1)
printf("-%c",b[k]);
flag = 1;
}
else
{
if(a[k] > 0)
printf("+");
if(a[k] != 1 && a[k] != -1)
printf("%d%c",a[k],b[k]);
else if(a[k] == 1)
printf("%c",b[k]);
else if(a[k] == -1)
printf("-%c",b[k]);
flag = 1;
}
}
if(a[10])
{
if(a[10] > 0 && flag)
printf("+");
printf("%d",a[10]);
flag = 1;
}
if(!flag)//没有答案
printf("0");
printf("\n");
}
return 0;
}
/*
99
0 0 0 0 0 0 0 0 0 -1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
-1 0 0 0 0 0 0 0 0 0
-1 -1 -1 -41 -1 -1 -1 -1 -1 -1
-1 5 -2 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1
0 0 0 0 0 -1 -1 -1 -1 -1
0 0 0 0 0 1 1 1 1 1
1 1 1 1 1 0 0 0 0 0
-1 -1 -1 -1 -1 0 0 0 0 0
1 1 1 1 1 1 1 1 1 0
*/
版权声明：本文博客原创文章。博客，未经同意，不得转载
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/4723404.html，如需转载请自行联系原作者
