pat_1002

源地址：http://www.patest.cn/contests/pat-a-practise/1002

就是一个多项式相加的过程，模拟一下即可。

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<set>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-3
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
int n,m;
#define Mod 1000000007
#define N 510
#define M 1000100
const int size = 10010;
const int mod = 9901;
char ch[M];
int key[30];
struct Pol{
	int exp;
	double coe;
	bool operator < (const Pol &x) const{
		return exp>x.exp;
	}
};
Pol a[15];
Pol b[15];
Pol c[N];
int vis[15];
int main(){
    while(scanf("%d",&n)!=EOF){
    	memset(vis,0,sizeof vis);
    	for(int i=0;i<n;i++){
    		scanf("%d%lf",&a[i].exp,&a[i].coe);
    	}
    	scanf("%d",&m);
    	for(int i=0;i<m;i++){
    		scanf("%d%lf",&b[i].exp,&b[i].coe);
    	}
    	int s = 0;
    	for(int i=0;i<n;i++){
    		int exp = a[i].exp;
    		double coe = a[i].coe;
    		for(int j=0;j<m;j++){
    			if(b[j].exp == exp){
    				coe += b[j].coe;
    				vis[j] = 1;
    				break;
    			}
    		}
    		c[s++] = (Pol){exp,coe};
    	}
    	for(int i=0;i<m;i++){
    		if(!vis[i]){
    			c[s++] = b[i];
    		}
    	}
    	sort(c,c+s);
    	int ans = 0;
    	for(int i=0;i<s;i++){
    		if(c[i].coe != 0)
    			ans++;
    	}
    	printf("%d",ans);
    	for(int i=0;i<s;i++){
    		if(c[i].coe == 0) continue;
    		printf(" %d %.1lf",c[i].exp,c[i].coe);
    	}
    	printf("\n");
    }
return 0;
}