hdu 5095 Linearization of the kernel functions in SVM（模拟，分类清楚就行）

题意：

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.

 

样例：

2
0 46 3 4 -5 -22 -8 -32 24 27                         --->                        46q+3r+4u-5v-22w-8x-32y+24z+27

2 31 -5 0 0 12 0 0 -49 12                             --->                        2p+31q-5r+12w-49z+12

 

代码：

char b[15]={'p','q','r','u','v','w','x','y','z'}; int a[15]; int T;int main(){//freopen("test.in","r",stdin);cin>>T;while(T--){rep(i,0,9) scanf("%d",&a[i]);int c=-1;rep(i,0,8) if(a[i]!=0) {c=i; break;}if(c==-1){printf("%d\n",a[9]);continue;}if(abs(a[c])==1)if(a[c]>0) printf("%c",b[c]); else printf("-%c",b[c]);elseprintf("%d%c",a[c],b[c]); ++c;rep(i,c,8){if(a[i]==0) continue;if(abs(a[i])==1)if(a[i]==1) printf("+%c",b[i]); else printf("-%c",b[i]);elseif(a[i]>0) printf("+%d%c",a[i],b[i]); else printf("%d%c",a[i],b[i]);}if(a[9]!=0){if(a[9]>0) printf("+%d\n",a[9]); else printf("%d\n",a[9]);}elseprintf("\n");}//fclose(stdin); }

 

转载于:https://www.cnblogs.com/fish7/p/4085144.html

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