UVA寫題記錄

UVA 1451

Description

求一個長度為N的01串長度至少為L的子串的數字平均值最大的起點和終點

Solution

可以看成斜率來做

但我寫的二分，二分平均值然後掃一遍就行了

O(nlogn)

#include<bits/stdc++.h>

using namespace std;

inline int read()
{
    int f = 1,x = 0;
    char ch;
    do
    {
        ch = getchar();
        if(ch == '-') f = -1;
    }while(ch < '
 
0'|| ch > '9');
    do
    {
        x = (x<<3) + (x<<1) + ch - '0';
        ch = getchar();
    }while(ch >= '0'&&ch <= '9');
    return f*x;
} 

#define eps 1e-6
const int MAXN = 100000 + 10;

int T,pos;
int ans1,ans2;
int n,L;
char a[MAXN];
double sum[MAXN];

inline bool check(double
 
 x)
{
    for(int i=1;i<=n;i++) sum[i] = sum[i-1] + a[i-1] - '0' - x;
    double minn = 0,maxx = -200000.0;
    for(int i=L;i<=n;i++)
    {
        if(minn >= sum[i-L]) minn = sum[i-L],pos=i-L+1;
        if(sum[i]-minn>maxx + eps)
        {
            maxx = sum[i] - minn;
            ans1 = pos;
            ans2 
 
= i;
        }
    }
    if(maxx + eps > 0) return 1;
    return 0;
}

int main()
{
    T = read();
    while(T--)
    {
        n = read(),L = read();
        scanf("%s",a);
        double l = 0.0 ,r = 1.0;
        for(int i=1;i<=100;i++)
        {
            double mid = (r+l)/2.0;
            if(check(mid)) l = mid;
            else r = mid;
        }
    //    check(l+eps);
        printf("%d %d\n",ans1,ans2);
    }
}