51nod1305 Pairwise Sum and Divide

1305 Pairwise Sum and Divide题目来源： HackerRank基准时间限制：1 秒 空间限制：131072 KB 分值: 5 难度：1级算法题 收藏  关注有这样一段程序，fun会对整数数组A进行求值，其中Floor表示向下取整：fun(A)    sum = 0    for i = 1 to A.length        for j = i+1 to A.length            sum = sum + Floor((A[i]+A[j])/(A[i]*A[j]))     return sum给出数组A，由你来计算fun(A)的结果。例如：A = {1, 4, 1}，fun(A) = [5/4] + [2/1] + [5/4] = 1 + 2 + 1 = 4。Input第1行：1个数N，表示数组A的长度(1 <= N <= 100000)。第2 - N + 1行：每行1个数A[i]（1 <= A[i] <= 10^9)。Output输出fun(A)的计算结果。Input示例31 4 1Output示例4

其实所有结果只要0,1,2.只需要要排序一遍,对1,2进行讨论即可。其实O(n)就可以做。

#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cctype>#include<cmath>#include<ctime>#include<string>#include<stack>#include<deque>#include<queue>#include<list>#include<set>#include<map>#include<cstdio>#include<limits.h>#define MOD 1000000007#define fir first#define sec second#define fin freopen("/home/ostreambaba/文档/input.txt", "r", stdin)#define fout freopen("/home/ostreambaba/文档/output.txt", "w", stdout)#define mes(x, m) memset(x, m, sizeof(x))#define Pii pair<int, int>#define Pll pair<ll, ll>#define INF 1e9+7#define inf 0x3f3f3f3f#define Pi 4.0*atan(1.0)#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long ll;typedef unsigned long long ull;const double eps = 1e-9;const int maxn = 100000+10;using namespace std;inline int read(){    int x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();    return x*f;}int num[maxn];int main(){    //freopen("/home/ostreambaba/文档/input.txt", "r", stdin);    //freopen("/home/ostreambaba/文档/output.txt", "w", stdout);    int n=read();    for(int i=0;i<n;++i){        num[i]=read();    }    int sum=0;    sort(num,num+n);    for(int i=0;i<n;++i){        if(num[i]==1){            int t=upper_bound(num,num+n,num[i])-num;            sum+=(n-t);            sum+=(t-1-i)*2;        }        if(num[i]==2){            int t=upper_bound(num,num+n,num[i])-num;            sum+=(t-1-i);        }    }    cout<<sum<<endl;    return 0;}