Input format:
The input is divided into 2 lines, and each line first gives the number of non-zero terms of the polynomial, and then enters a polynomial non-zero term coefficient and exponent in an exponentially descending manner (the absolute value is an integer not exceeding 1000). Numbers are separated by spaces.
Output format:
The output is divided into 2 lines, and the coefficients and exponents of the product polynomial and the non-zero term of the sum polynomial are output in an exponentially descending manner, respectively. Numbers are separated by spaces, but there can be no extra spaces at the end. A zero polynomial should output 0 0.
Input sample:
4 3 4 -5 2 6 1 -2 0
3 5 20 -7 4 3 1
Sample output:
15 24 -25 22 30 21 -10 20 -21 8 35 6 -33 5 14 4 -15 3 18 2 -6 1
5 20 -4 4 -5 2 9 1 -2 0
代码长度限制16 KB
时间限制200 ms
内存限制64 MB
Answer:
#include<stdio.h>
#define N 10000
int main() {
int a[N]= {
0};
int b[N]= {
0};
int c[N]= {
0};
int d[N]= {
0};
int i,m,f;
scanf("%d",&i);
while(i--) {
scanf("%d %d",&m,&f);
a[f]+=m;
}
scanf("%d",&i);
while(i--) {
scanf("%d %d",&m,&f);
b[f]+=m;
}
for(int i=N-1; i>=0; i--) {
if(a[i]) {
for(int j=0; j<N; j++) {
if(b[j]) {
c[i+j]+=a[i]*b[j];
}
}
}
}
int cnt=0;
for(int i=N-1; i>=0; i--) {
if(c[i]) {
if(cnt)printf(" ");
printf("%d %d",c[i],i);
cnt++;
}
}
if(!cnt)printf("0 0");
for(int i=N-1; i>=0; i--)
if(a[i])
d[i]+=a[i];
for(int j=0; j<N; j++)
if(b[j])
d[j]+=b[j];
printf("\n");
cnt=0;
for(int i=N-1; i>=0; i--) {
if(d[i]) {
if(cnt)printf(" ");
printf("%d %d",d[i],i);
cnt++;
}
}
if(!cnt)printf("0 0");
return 0;
}
test:
