HDU 5095--Linearization of the kernel functions in SVM【模拟】
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
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.
j of the function f(x,y,z) = ax^2 + by^2 + cy^2 + dxy + eyz + fzx + gx + hy + iz + j.
2
0 46 3 4 -5 -22 -8 -32 24 27
2 31 -5 0 0 12 0 0 -49 12
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;
}
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