程序设计基础41 tips 关于数组从后往前遍历

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1009 Product of Polynomials （25 分）

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 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 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:

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1
2
3
2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

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1
2
3 3 3.6 2 6.0 1 1.6

一，注意点

1，开的数组长度是max_n，应当从max_n-1开始遍历到下标为0的，不能从max_n开始。

二，代码

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#include<cstdio>
using namespace std;
const int max_n =2020;
double arr[max_n]={0};
struct Polynomial{
int ex;
double coe;
}pol[13];
int main(){
int num_1=0;
int num_2=0;
int K=0;
double Nk=0;
scanf("%d",&num_1);
for(int i=0;i<num_1;i++){
scanf("%d %lf",&pol[i].ex,&pol[i].coe);
}
scanf("%d",&num_2);
for(int i=0;i<num_2;i++){
int a=0;
double b=0;
scanf("%d %lf",&K,&Nk);
for(int j=0;j<num_1;j++){
a=K+pol[j].ex;
b=Nk*pol[j].coe;
arr[a]+=b;
}
}
int ans=0;
/************************************
************************************
从max_n-1开始遍历*********************/
for(int i=max_n-1;i>=0;i--){
if(arr[i]!=0){
ans++;
}
}
printf("%d",ans);
for(int i=max_n-1;i>=0;i--){
if(arr[i]!=0){
printf(" %d %.1f",i,arr[i]);
}
}
return 0;
}