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 ",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 ",a[9]); else printf("%d ",a[9]); } else printf(" "); } //fclose(stdin); }