JZOI1169A 平均数Ave

NOI算法解析

最新推荐文章于 2024-09-19 08:40:39 发布

转载最新推荐文章于 2024-09-19 08:40:39 发布 · 134 阅读

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原文链接：http://www.cnblogs.com/ARZhu-NOIpAK/p/6862039.html

本文介绍了一个用于解决特定信息学竞赛问题的算法实现。通过自定义输入读取函数提高效率，并利用归并排序进行数列排序及逆序对计数，以此来辅助查找满足条件的中位数，从而确定最优解。

#include <cstdio>
#include <cmath>
#define lztin() read()
#define ztyout( a ) printf( "%.4lf", a )
#define eps 1e-5
#define ll long long
using namespace std;
inline int read() {
    int x = 0,tmp = 1;char ch = getchar();
    while( ch < '0' || ch > '9' ) {if ( ch == '-' ) tmp = -1; ch = getchar();}
    while( ch >= '0' && ch <= '9'){x = x * 10 + ch - '0'; ch = getchar(); }
    return x * tmp;
}
int A[110000], N, K, cnt;
double B[1100000], T[1100000];
double ans;
void msort( int l, int r ) {
    if( l == r ) return;
    int mid = l + r >> 1;
    msort( l, mid ); 
    msort( mid + 1, r );
    int i = l, k = l, j = mid + 1;
    while( i <= mid && j <= r ) {
        if( B[i] > B[j] ) {
            cnt += mid - i + 1;
            T[k] = B[j];
            k ++, j ++;
        } else {
            T[k] = B[i];
            k ++, i ++;
        }
    }
    while( i <= mid ) {
        T[k] = B[i];
        k ++, i ++;
    }
    while( j <= r ) {
        T[k] = B[j];
        k ++, j ++;
    }
    for( int w = l ; w <= r ; ++ w ) B[w] = T[w];
}
int check( double mid ) {
    B[0] = 0; cnt = 0;
    for( int i = 1 ; i <= N ; ++ i ) {
        B[i] = B[i - 1] + (double)A[i] - mid;
    }
    msort( 1, N );
    for( int i = 1 ; i <= N ; ++ i )
        cnt += ( B[i] < 0 );
    return ( cnt >= K );
}
int main() {
    N = lztin(), K = lztin();
    for( int i = 1 ; i <= N ; ++ i ) {
        A[i] = lztin();
    }
    double l = 0, r = (double)1e9, mid;
    while( fabs( l - r ) > eps ) {
        mid = ( l + r ) / 2.0;
        if( check( mid ) ) ans = mid, r = mid;
        else l = mid;
    }
    ztyout( ans );
    
    return 0;
}

转载于:https://www.cnblogs.com/ARZhu-NOIpAK/p/6862039.html