pat1002

1002. A+B for Polynomials (25)

時間限制

400 ms

內存限制

65536 kB

代碼長度限制

16000 B

判題程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

#include<cstdio>
#include<vector>
#include<iostream>
#include<cstring>
#include<algorithm>
const int maxn = 1005;
using namespace std;
vector<double> pile[maxn];
int flag[maxn];
int main(){
	int k, a, n=0;
	double b,sum;
	int N[12],c=0;
	memset(flag,0,sizeof(flag));
	for(int j =0; j < 2; j++){
		scanf("%d",&k);
		for(int i =0; i < k; i++){
			scanf("%d %lf", &a, &b);
			pile[a].push_back(b);
			if(flag[a]==0){
				flag[a]=1;
				N[c++]=a;
			}
		}
	}
	sort(N,N+c);
	printf("%d ", c);
	for(int i =c-1; i > 0; i--){
		sum = 0.0;
		for(int j = 0 ; j < pile[N[i]].size(); j++){
			sum += pile[N[i]][j];
		}
		
		printf("%d %0.1lf ", N[i], sum);
	}
	sum=0.0;
	for(int j = 0 ; j < pile[N[0]].size(); j++){
		sum += pile[N[0]][j];
		printf("%d %0.1lf", N[0], sum);
	}
	return 0;
}

運行結果 有幾個測試點未通過 忽略了多項式相加係數爲0的情況

然後將代碼改了一下，上面的空格輸出麻煩了，改了一下，結果如下 之後又看了一下別人的 發現用結構體很好 寫的時候就想到了vector容器 導致有點麻煩了

#include<cstdio>
#include<vector>
#include<iostream>
#include<cstring>
#include<algorithm>
const int maxn = 1005;
using namespace std;
vector<double> pile[maxn];
int flag[maxn];
int main(){
	int k, a, n=0;
	double b,sum;
	int N[12],c=0;
	memset(flag,0,sizeof(flag));
	for(int j =0; j < 2; j++){
		scanf("%d",&k);
		for(int i =0; i < k; i++){
			scanf("%d %lf", &a, &b);
			pile[a].push_back(b);
			if(flag[a]==0){
				flag[a]=1;
				N[c++]=a;
			}
		}
	}
	sort(N,N+c);
	int m=c;
	//printf("%d", c);
	for(int i =c-1; i >= 0; i--){
		sum = 0.0;
		for(int j = 0 ; j < pile[N[i]].size(); j++){
			sum += pile[N[i]][j];
		}
		if(sum==0){
			m--;
		}
	}
	printf("%d", m);
	for(int i =c-1; i >= 0; i--){
		sum = 0.0;
		for(int j = 0 ; j < pile[N[i]].size(); j++){
			sum += pile[N[i]][j];
		}
		if(sum!=0)
		printf(" %d %0.1lf", N[i], sum);
	}
	return 0;
}