1002 A+B for 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 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《算法笔记》
#include<cstdio>
const int max_n=1111;
double p[max_n]={};
int main(){
int k,n,count=0;
double a;
scanf("%d",&k);
for(int i=0;i<k;i++){
scanf("%d %lf",&n,&a);
p[n]+=a;
}
scanf("%d",&k);
for(int i=0;i<k;i++){
scanf("%d %lf",&n,&a);
p[n]+=a;
}
for(int i=0;i<max_n;i++){
if(p[i]!=0){
count++;
}
}
printf("%d",count);
for(int i=max_n-1;i>=0;i--){
if(p[i]!=0)printf(" %d %.1f",i,p[i]);
}
return 0;
}
ac：
#include<cstdio>
#include<cmath>
const int maxn=1001;
int main(){
float A[maxn],B[maxn],C[maxn];
for(int i=0;i<maxn;i++){
A[i]=10001;
B[i]=10001;
C[i]=10001;
}
int a, b,temp;
scanf("%d",&a);
for(int i=0;i<a;i++){
scanf("%d",&temp);
scanf("%f",&A[temp]);
}
scanf("%d",&b);
for(int i=0;i<b;i++){
scanf("%d",&temp);
scanf("%f",&B[temp]);
}
int H=0;
for(int i=0;i<maxn;i++){
if(10001!=(int)A[i]&&10001!=(int)B[i]){
C[i]=B[i]+A[i];
if((int)C[i]!=0)H++;
}
else if(10001!=(int)A[i]||10001!=(int)B[i]){
if((int)A[i]!=10001)C[i]=A[i];
else C[i]=B[i];
if((int)C[i]!=0)H++;
}
}
printf("%d",H);
for(int i=maxn-1;i>=0;i--){
if((int)C[i]!=10001&&(int)C[i]!=0){
printf(" %d %.1f",i,C[i]);
}
}
return 0; //float 大小比较问题~~~~~~~~！！！！！！！！！！ 且没考虑到消去为0的情况
}//好久前写的代码了2333，仅做复时使用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:
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《算法笔记》：
#include<cstdio>
struct Poly{
int exp;//指数
double cof;//系数
}poly[1001];//第一个多项式
double ans[2001];//存放结果
int main(){
int n,m,number=0;
scanf("%d",&n); //第一个多项式中非零系数的项数
for(int i=0;i<n;i++){
scanf("%d %lf",&poly[i].exp,&poly[i].cof);
}
scanf("%d",&m);//第二个多项式中非零系数的项数
for(int i=0;i<m;i++){
int exp;
double cof;
scanf("%d %lf",&exp,&cof);//第二个多项式的指数和系数
for(int j=0;j<n;j++){//与第一个多项式中的每一项相乘
ans[exp+poly[j].exp]+=(cof*poly[j].cof);
}
}
for(int i=0;i<=2000;i++){
if(ans[i]!=0.0)number++;
}
printf("%d",number);
for(int i=2000;i>=0;i--){
if(ans[i]!=0.0){
printf(" %d %.1f",i,ans[i]);
}
}
return 0;
}
ac：
#include<cstdio>
#include<vector>
#include<algorithm>
#include<map>
using namespace std;
int k,ex,en=0;
double cf;
struct Node{
int e;
double c;
};
vector<Node> a1,a2,ans;
map<int,int> p;
bool cmp(Node a,Node b){
return a.e>b.e;
}
int main(){
scanf("%d",&k);
for(int i=0;i<k;i++){
scanf("%d %lf",&ex,&cf);
a1.push_back(Node{ex,cf});
}
sort(a1.begin(),a1.end(),cmp);
scanf("%d",&k);
for(int i=0;i<k;i++){
scanf("%d %lf",&ex,&cf);
a2.push_back(Node{ex,cf});
}
sort(a2.begin(),a2.end(),cmp);
for(int i=0;i<a1.size();i++){
for(int j=0;j<a2.size();j++){
ex=a1[i].e+a2[j].e;
cf=a1[i].c*a2[j].c;
if(p.find(ex)==p.end()){
ans.push_back(Node{ex,cf});
p[ex]=en;
en++;
}
else{
int temp=p[ex];
ans[temp].c+=cf;
}
}
}
sort(ans.begin(),ans.end(),cmp);
for(vector<Node>::iterator it=ans.begin();it!=ans.end();it++){
if((int)(*it).c==0)ans.erase(it);
}
printf("%d",ans.size());
for(int i=0;i<ans.size();i++){
if((int)ans[i].c!=0)printf(" %d %.1lf",ans[i].e,ans[i].c);
}
return 0;
}