1009.PRoduct of Polynomials (25)
pat-al-1009
2017-02-15
哈希表的應(yīng)用坑見注釋/** * pat-al-1009 * 2017-02-15 * C version * Author: fengLian_s */#include<stdio.h>int main(){ double hs1[1001] = {0}, hs2[1001], result[2001] = {0};//坑：要記得給hs1初始化，不然會不知道數(shù)組里存的到底是什么鬼 freopen("in.txt", "r", stdin); int k1, k2; scanf("%d", &k1); for(int i = 0;i < k1;i++) { int tmpE; double tmpC; scanf("%d%lf", &tmpE, &tmpC); //printf("tmpE = %d, tmpC = %.1lf/n", tmpE, tmpC); hs1[tmpE] = tmpC; } scanf("%d", &k2); for(int i = 0;i < k2;i++) { int tmpE; double tmpC; scanf("%d%lf", &tmpE, &tmpC); //printf("tmpE = %d, tmpC = %.1lf/n", tmpE, tmpC); for(int j = 0;j <= 1000;j++) { if(hs1[j]*tmpC != 0) { //printf("hs1[%d] = %.1lf, tmpC = %.1lf/n", j, hs1[j], tmpC); result[tmpE+j] += hs1[j]*tmpC; } } hs2[tmpE] = tmpC; } int cnt = 0; for(int i = 0;i <= 2000;i++) { if(result[i] != 0) cnt++; } printf("%d", cnt); for(int i = 2000;i >= 0;i--) { if(result[i] != 0) printf(" %d %.1lf", i, result[i]); } putchar('/n');}-FIN-
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