HDU 5095--Linearization of the kernel functions in SVM【模拟】

最新推荐文章于 2018-05-16 19:36:48 发布

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最新推荐文章于 2018-05-16 19:36:48 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/jhcelue/p/7403888.html

Linearization of the kernel functions in SVM

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2232 Accepted Submission(s): 598

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

最烦做的就是模拟题，恶心啊。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <iostream>
using namespace std;

char str[] = {'p', 'q', 'r', 'u', 'v', 'w', 'x', 'y', 'z', '\0'};
int a[15];

int main (){
    int T;
    scanf("%d", &T);
    while(T--){
        for(int i = 0; i < 10; ++i)
            scanf("%d", &a[i]);
        int ans = 0;
        int flag = 0;
        for(int i = 0; i < 9; ++i){

            if(a[i] == 0) continue;
            if(a[i] == 1 ){
                if(flag)
                printf("+");
                else
                    flag++;
                printf("%c", str[i]);
                continue;
            }
            if(a[i] == -1){
                printf("-%c", str[i]);
                flag++;
                continue;
            }
            if(a[i] > 0){
                if(flag)
                    printf("+");
                else
                    flag++;
                printf("%d%c", a[i], str[i]);
            }
            else
                printf("%d%c", a[i], str[i]), flag++;
            }
            if(a[9] > 0){
                if(flag)
                    printf("+");
                else
                    flag++;
                printf("%d", a[9]);
            }
            else if( a[9] < 0){
                flag ++;
                printf("%d", a[9]);
            }
            if(!flag)
                printf("0");
            printf("\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/jhcelue/p/7403888.html