PTA 1048 Find Coins

题目描述

分析：

解法1：利用散列表记录数字的个数即可，判断i和m-i是否都存在，注意假如i==m-i则要判断i出现的次数是否大于等于2

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int HashTable[1005];
int main(){
   int n,m;
   cin>>n>>m;
   int a;
   while(n--){
   	cin>>a;
   	HashTable[a]++;
   }
   for(int i=0;i<1005;i++){
   	if(HashTable[i]&&HashTable[m-i]){
   		if(i==m-i&&HashTable[i]<=1)
   		continue;
   		cout<<i<<" "<<m-i<<endl;
   		return 0;
	   }
   }
   cout<<"No Solution"<<endl;
   return 0;
}

解法2：二分查找

#include<cstdio>
#include<algorithm>
using namespace std;
int a[100005];
int binarySearch(int l,int r,int key){
	int mid;
	while(l<=r){
		mid=(l+r)/2;
		if(a[mid]==key){
			return mid;
		}
		else if(a[mid]>key) {
			r=mid-1;
		}
		else{
			l=mid+1;
		}
	}
	return -1;
} 
int main(){
   int n,m,i;
   scanf("%d %d",&n,&m);
   for(i=0;i<n;i++){
   	scanf("%d",&a[i]);
   }
   sort(a,a+n);
   for(i=0;i<n;i++){
   	int pos=binarySearch(0,n-1,m-a[i]);
   	if(pos!=-1&&i!=pos){
   		printf("%d %d",a[i],m-a[i]);
   		break;
	   }
   }
   if(i==n)
   printf("No Solution\n");
   return 0;
}

解法3：双指针法

#include<cstdio>
#include<algorithm>
using namespace std;
int a[100005];
void twoPointers(int n,int m){
	int i=0;
	int j=n-1;
	while(i<j){
		if(a[i]+a[j]==m) break;
		else if(a[i]+a[j]<m){
			i++;
		}
		else{
			j--;
		}
	}
	if(i<j){
		printf("%d %d\n",a[i],a[j]);
	}
	else{
		printf("No Solution\n");
	}
}
int main(){
   int n,m,i;
   scanf("%d %d",&n,&m);
   for(i=0;i<n;i++){
   	scanf("%d",&a[i]);
   }
   sort(a,a+n);
   twoPointers(n,m);
   return 0;
}