PAT A 1002 A+B for Polynomials

题目链接

一刷

• 一定要注意理解题目意思啊，人家要输出非零项的个数和非零项，注意正负相消
• 然后就是变量范围，It is given that 1≤K≤10，0≤N**K<⋯<N2<N1≤1000.K是每行最多给10项，N是指每一项，数组要根据这个来定义，所以要大于1000
• 最后要注意输出的空格啊！！！

#include <iostream>
#include <vector>
using namespace std;

vector<double> nums(1100, 0.0);

int main() {
	int k, n = 0;
	double m = 0.0;

	while (cin>>k) {
		for (int i = 0; i < k; i++) {
			cin >> n >> m;
			nums[n] += m;
		}
	}
	int len = nums.size()-1;
	int cnt = 0;
	for (int i = 0; i < len; i++) {
		if (nums[i] != 0) {
			cnt++;
		}
	}
	printf("%d",cnt);
	for (int i = len; i >=0; i--) {
		if (nums[i] == 0.0) continue;
		printf(" %d %0.1f", i, nums[i]);
	}
	return 0;
}

二刷

#include <cstdio>
const int maxn=1010;
double poly[maxn];
int main(){
	int n,e;
	double c;
	scanf("%d",&n);
	for(int i=0;i<maxn;i++) poly[i]=0;
	for(int i=0;i<n;i++){
		scanf("%d%lf",&e,&c);
		poly[e]=c;
	}
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d%lf",&e,&c);
		poly[e]+=c;
	}
	int cnt=0;
	for(int i=1000;i>=0;i--){
		if(poly[i]!=0){
			cnt++;
		}
	}
	printf("%d",cnt);
	for(int i=1000;i>=0;i--){
		if(poly[i]!=0){
			printf(" %d %.1f",i,poly[i]);
		}
	}
	
	return 0;
}