1002. A+B for Polynomials (25)
1002. A+B for Polynomials (25) This time, you are supposed to find A+B where A and B are two polynomials. Input 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 For each test case you should output the sum 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 to 1 decimal place. Sample Input 2 1 2.4 0 3.2 2 2 1.5 1 0.5 Sample Output 3 2 1.5 1 2.9 0 3.2
分析:题目是两个多项式相加。有坑点在第一次提交的时候没注意,只拿到了17分。由于采用了指数和系数的表示形式,那么当两个多项式相加的时候,就有可能某一项的系数为0,那么这一项就不用输出了。在做题目的时候想得还是不够全面。
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> using namespace std; int main() { double ans[1010]; int flag[1010]; memset(flag,0,sizeof(flag)); int cnt=0; for(int input=0;input<2;input++) { int k; scanf("%d",&k); { for(int i=0;i<k;i++) { int index; double num; cin>>index>>num; if(flag[index]==0) { flag[index]=1; cnt++; ans[index]=num; } else { ans[index]+=num; //判断系数是否为0,若为0去掉这一项 if(ans[index]<0.0000000001) { flag[index]=0; cnt--; } } } } } cout<<cnt; int j=1010-1; for(;j>=0;j--) { if(flag[j]==1) { printf(" %d %.1lf",j,ans[j]); } } cout<<endl; return 0; }
——来自 熊猫 [http://www.cnblogs.com/xiongmao-cpp/]