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  • 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
     
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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;
    }
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  • 原文地址:https://www.cnblogs.com/w-y-1/p/5750311.html
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