立志用最少的代碼做最高效的表達(dá)

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PAT甲級(jí)最優(yōu)題解——>傳送門

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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

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測(cè)試點(diǎn)0：多項(xiàng)式相乘，可能出現(xiàn)系數(shù)為0的情況，因此，最后輸出的項(xiàng)數(shù)可能計(jì)算錯(cuò)誤。

解法：數(shù)組硬解， 其實(shí)換成結(jié)構(gòu)體存儲(chǔ)效率會(huì)高一點(diǎn)，但考慮到最大數(shù)據(jù)量只有1k，因此也無所謂了。

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#include<bits/stdc++.h> using namespace std; typedef long long gg;gg a[1010] = {0}, b[1010] = {0}, c[2020] = {0}; int main() {gg k,x; cin >> k; while(k--) {cin >> x; double d; cin >> d;a[x] = (gg)(d*100);}cin >> k; while(k--) {cin >> x; double d; cin >> d;b[x] = (gg)(d*100);}gg num = 0;for(gg i = 0; i <= 1000; i++) for(gg j = 0; j <= 1000; j++) {if(a[i] == 0) continue;if(b[j] != 0) c[i+j] += a[i]*b[j]; }for(int i = 0; i <= 2000; i++) if(c[i] != 0) num++;printf("%d", num);for(int i = 2000; i >= 0; i--) if(c[i] != 0) printf(" %d %.1lf", i, c[i]/10000.0);return 0; }

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