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]);

        else

            printf("%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]);

            else

                if(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]);

        }

        else

            printf("\n");

    }

    //fclose(stdin);

}

巴特西

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

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