This time, you are supposed to find A*B where A and B are two polynomials.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.
Output Specification:
For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.
Sample Input 2 1 2.4 0 3.2 2 2 1.5 1 0.5 Sample Output 3 3 3.6 2 6.0 1 1.6参考代码1：
#include<cstdio> const int N=1001; const int RN=1000001; double p1[N]={ },p2[N]={ },pro[RN]={ }; int main(){ int k,count1[N]={ }; scanf("%d",&k); int izero=0; for(int j=0;j<k;j++){ int n; double an; scanf("%d%lf",&n,&an); p1[n]=an; count1[izero]=n; izero++; } scanf("%d",&k); for(int j=0;j<k;j++){ int n; double an; scanf("%d%lf",&n,&an); p2[n]=an; for(int i=0;i<izero;i++){ pro[n+count1[i]]+=p2[n]*p1[count1[i]]; } } int count=0; for(int i=0;i<RN;i++){ if(pro[i]!=0){ count++; } } if(count!=0) printf("%d",count); else{ printf("0\n"); return 0; } for(int i=1000000;i>-1;i--){ if(pro[i]!=0){ printf(" %d %.1f",i,pro[i]); } } printf("\n"); return 0; }参考代码2：
#include<cstdio> struct { int exp; double cof; }poly[1001]; int main(){ double ans[2001]={ }; int k1; scanf("%d",&k1); for(int i=0;i<k1;i++){ scanf("%d %lf",&poly[i].exp,&poly[i].cof); } int k2; scanf("%d",&k2); for(int i=0;i<k2;i++){ int n; double an; scanf("%d %lf",&n,&an); for(int i=0;i<k1;i++){ ans[n+poly[i].exp]+=an*poly[i].cof; } } int count=0; for(int i=0;i<2001;i++){ if(ans[i]!=0) count++; } printf("%d",count); for(int i=2000;i>-1;i--){ if(ans[i]!=0){ printf(" %d %.1f",i,ans[i]); } } printf("\n"); return 0; }