題干：

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?N?1???a?N?1?????N?2???a?N?2?????...?N?K???a?N?K????

where?K?is the number of nonzero terms in the polynomial,?N?i???and?a?N?i?????(i=1,2,?,K) are the exponents and coefficients, respectively. It is given that?1≤K≤10,?0≤N?K??<?<N?2??<N?1??≤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

?題目大意：

按照輸入的格式 給定兩個多項式A和B，讓你按照相同的格式輸出兩個多項式的乘積，也就是A×B。

解題報告：

就是個模擬題，但是注意細節不少。

1.首先觀察輸入可以發現當系數為0的時候這一項是需要省略的。

2.其次最后輸出的第一個數不是最高次項，而是后面需要跟著的數的個數。

3.次數需要從高往低輸出。

4.注意進行多項式乘積的時候，跟輸入的兩個K是沒有關系的，而是跟兩組N1...Nk是有關系的，又因為這里保證了N是由大到小的，也就是只和兩個N1是有關系的。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; double A[MAX],B[MAX],C[MAX],b; int n,m,mx,a; int main() {cin>>n;for(int i = 1; i<=n; i++) {scanf("%d%lf",&a,&b);A[a] = b;if(i == 1) mx = a;}cin>>m;for(int i = 1; i<=m; i++) {scanf("%d%lf",&a,&b);B[a] = b;if(i == 1) mx += a;}for(int i = 0; i<=mx; i++) {for(int j = 0; j<=mx; j++) {C[i+j] += A[i]*B[j];}}int ans = 0;for(int i = 0; i<=mx; i++) {if(C[i] != 0) ans++;}printf("%d",ans);for(int i = mx; i>=0; i--) {if(C[i] == 0) continue;printf(" %d %.1f",i,C[i]);}return 0 ; }

?

總結

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