枚舉 練習

題目來源:

http://coder.buct.edu.cn/oj/Contest/Contest.aspx?cid=1121

第一題就是簡單的求次大數。

第二題需要注意本身與本身也可以構成一個序對。

第三題儘量避免使用浮點型。

第四題注意時間控制。

 

1:
#include<stdio.h>
int main()
{
	int n,i,t1,t2,t;
	while(scanf("%d",&n)!=EOF){
		scanf("%d %d",&t1,&t2);
		if(t1<t2){
			t1=t1^t2;
			t2=t1^t2;
			t1=t1^t2;
		}//t1保存最大，t2保存次大
		else if(t1 == t2){
			t2 = -1000001 ;
		}
		for(i = 2;i < n;i++){
			scanf("%d",&t);
			if(t>t1){
				t2 = t1;
				t1 = t;
			}
			else if(t>t2&&t<t1)
				t2 = t;
		}
		if(t2 == -1000001)
			printf("No such number\n");
		else
			printf("%d\n",t2);
	}
	return 0;
}
2:
#include<stdio.h>
const int maxn = 2001;
int list[maxn];
int main()
{
	int sum,k,i;
	while(scanf("%d%d",&sum,&k)!=EOF){
		for(i=0;i<k;i++)
			scanf("%d",&list[i]);
		int num = 0; 
		for(i=0;i<k;i++){
			for(int j=i;j<k;j++)//注意本身和本身也是一個數對
				if(list[i] + list[j] == sum)
					num++;
		}
		printf("%d\n",num);
	}
	return 0;
}
3：
#include<stdio.h>
int a,b;
typedef long long LL;
int gcd(int a,int b){
	return b?gcd(b,a%b):a;
}
LL abs(LL x)
{
	return x<0?-x:x;
}
bool little(int a1,int b1,int a2,int b2)
{
	return abs ((LL)a1*b*b2 - (LL)a*b1*b2)
		 < abs ((LL)a2*b*b1 - (LL)a*b1*b2);
}
int main()
{
	while(scanf("%d %d",&a,&b)!=EOF){
		int c=1,d=1;
		for(int i = 1;i<= 32767;i++){
			int lf = 1,rt = 32767,mid;
			while(lf+1 < rt){
				mid = lf + rt >> 1;
				little(lf,i,rt,i)?rt = mid:lf = mid;
			}
			mid = little(lf,i,rt,i)?lf:rt;
			int tmp = gcd(mid , i);
			if((mid/tmp == a)&&(i/tmp ==b)){
				if(mid == 1)
					mid++;
				else
					mid--;
			}
			if(little(mid,i,c,d)){
				c = mid;
				d = i;
			}
		}
		printf("%d %d\n",c,d);
	}
	return 0;
}
4:
#include<cstdio>
#include<map>
#include<iostream>
#include <cstdlib>
#include <set>
#include<algorithm>
using namespace std;
const int maxn = 2005;
struct node
{
	int x,y;
	friend bool operator < (node a,node b){
		if(a.x == b.x)
			return a.y < b.y;
		return a.x < b.x;
	}
}list[maxn];
set <node> S;
int yes(node a ,node b)
{
	node c1,d1;
	int num = 0;
	c1.x = a.x - (b.y - a.y);
	c1.y = a.y + (b.x - a.x);
	d1.x = b.x - (b.y - a.y);
	d1.y = b.y + (b.x - a.x);
	if(S.find(c1)!=S.end()&&S.find(d1)!=S.end()){
		num++;
	}
	return num;
}
int cmp(node a,node b)
{
	if(a.x == b.x)
		return a.y>b.y;
	return a.x>b.x;
}
int main()
{
	int n,i;
	while(scanf("%d",&n)!=EOF&&n){
		for(i=0;i<n;i++){
			scanf("%d %d",&list[i].x,&list[i].y);
			S.insert(list[i]);
		}
		sort(list,list+n,cmp);
		int num = 0;
		for(i = 0;i < n;i ++){
			for(int j = i+1;j<n;j ++){
					num+=yes(list[i],list[j]);
			}
		}
		printf("%d\n",num/2);
		S.clear();
	}
	return 0;
}