模拟 HDOJ 5095 Linearization of the kernel functions in SVM
/*
题意:表达式转换
模拟:题目不难,也好理解题意,就是有坑!具体的看测试样例。。。
*/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <ctime>
#include <cstdlib>
using namespace std; const int MAXN = 1e4 + ;
const int INF = 0x3f3f3f3f; char s[] = {'p', 'q', 'r', 'u', 'v', 'w', 'x', 'y', 'z'}; int main(void) //HDOJ 5095 Linearization of the kernel functions in SVM
{
//freopen ("F.in", "r", stdin); int t; scanf ("%d", &t);
while (t--)
{
int a[]; bool first = false; int j = -;
for (int i=; i<=; ++i) scanf ("%d", &a[i]);
for (int i=; i<=; ++i)
{
j++;
if (!a[i]) continue;
if (a[i] > && first) putchar ('+'); if (a[i] == -)
{
if (i < ) putchar ('-');
else printf ("-1");
}
else
{
if (a[i] == )
{
if (i == ) putchar ('');
}
else printf ("%d", a[i]);
} first = true;
if (j <= ) printf ("%c", s[j]);
}
if (!first) printf ("");
puts ("");
} return ;
} /*
21
0 46 3 4 -5 -22 -8 -32 24 27
2 31 -5 0 0 12 0 0 -49 12
0 1 0 0 0 0 0 0 0 -1
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
1 -1 1 -1 1 1 1 1 1 0
0 46 3 4 -5 -22 -8 -32 24 27
2 31 -5 0 0 12 0 0 -49 12
0 0 0 0 0 0 0 0 0 0
1 2 3 4 5 6 7 8 9 10
-1 2 3 -5 8 3 0 6 9 20
1 -1 1 1 1 -1 -1 -1 0 0
10000 123 123 123 456 12354 123 12345 45 10110
0 0 0 1 2 3 -1 -2 -3 56
0 0 0 -1 -2 0 0 1 23 45
*/ /*
46q+3r+4u-5v-22w-8x-32y+24z+27
2p+31q-5r+12w-49z+12
q-1
0
1
p-q+r-u+v+w+x+y+z
46q+3r+4u-5v-22w-8x-32y+24z+27
2p+31q-5r+12w-49z+12
0
p+2q+3r+4u+5v+6w+7x+8y+9z+10
-p+2q+3r-5u+8v+3w+6y+9z+20
p-q+r+u+v-w-x-y
10000p+123q+123r+123u+456v+12354w+123x+12345y+45z+10110
u+2v+3w-x-2y-3z+56
-u-2v+y+23z+45
-u-2v+y+23z+45
-u-2v+y+23z+45
-u-2v+y+23z+45
-u-2v+y+23z+45
-u-2v+y+23z+45
-u-2v+y+23z+45
*/
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