原題分類于排序 大部分人用二分查找&排序即可AC 用了vector和set 參考另一篇 Ananagrams UVa156的思想 在輸入 匹配上可以提高效率
#include <iostream>#include <cstdio>#include <sstream>#include <set>#include <bitset> #include <queue> #include <stack> #include <list>#include <vector>#include <map>#include <string>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef set<int> Set;typedef vector<int> Vec;typedef set<int>::iterator It;typedef long long ll;#define mem(s,n) memset(s,n,sizeof(s))int main(int argc, char *argv[]){ Set s;Vec v; int K,N,temp; scanf("%d%d",&K,&N); while(N--) { scanf("%d",&temp); if(!s.empty()&&s.find(temp)!=s.end()) { v.push_back(temp); v.push_back(K-temp); s.erase(temp);//匹配到則erase } else s.insert(K-temp); } sort(v.begin(),v.end()); if(!v.empty()) for(int i=0;i<v.size()/2;i++) 可以看出主要是sort和 讀取時(shí)處理 花了不少時(shí)間 還有另一種思路復(fù)雜度相似#include <iostream>#include <cstdio>#include <sstream>#include <set>#include <bitset> #include <queue> #include <stack> #include <list>#include <vector>#include <map>#include <string>#include <cstring>#include <cmath>#include <algorithm>using namespace std;#define ll long long#define mem(s,n) memset(s,n,sizeof(s))#define Set set<int> #define Vec vector<int>int main(int argc, char *argv[]){ int K,N,a[50010]; scanf("%d%d",&K,&N); for(int i=0;i<N;i++) { scanf("%d",&a[i]); if(a[i]>=K/2) a[i]=K-a[i];// } int bad=1,ans=0; sort(a,a+N); for(int i=0;i<N-1;i+=2) if(a[i]==a[i+1]) { printf("%d %d/n",a[i],K-a[i]); bad=0; } if(bad) printf("No solution/n"); return 0;}仍有更好的奇技淫巧
#include <stdio.h>#include <algorithm>using namespace std;const int N = 50005;int k, n;int a[N];int main() { int i, j; scanf("%d %d", &k, &n); for (i = 0; i < n; i++) { scanf("%d", &a[i]); } sort(a, a + n); bool flag = false; i = 0; j = n - 1; while (i < j) { n = a[i] + a[j]; if (n == k) { printf("%d %d/n", a[i++], a[j--]); flag = true; } else if (n < k) { i++; } else { j--; } } if (!flag) { puts("No Solution"); }}新聞熱點(diǎn)
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