Poj3061 subsequence (Binary prefix and method + ruler acquisition)

Binary + prefix and Method

The length of the sub-sequence meeting the condition is between (0, n, sum [x + I]-sum [I] is the sum of the elements whose sequence length is X starting from the I element. Prefix and time that can be counted in O (N)

The value of sum [I. Use the second part to find the minimum length of the sub-sequence that meets the condition.

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<queue>#include<set>#include<map>#include<vector>#include<cmath>#define ll __int64#define INF 0x3fffffffusing namespace std;int sum[100005];int a[100005];int n,s;bool C(int x){    bool flag=false;    for(int i=0;i<n-x;i++)    {        if(sum[x+i]-sum[i]>=s)        {            flag=true;            break;        }    }    if(flag) return true;    else return false;}int solve(){    int l=0,r=n+1;    while(r-l>1)    {        int mid=(l+r)/2;        if(C(mid)) r=mid;        else l=mid;    }    return r;}int main(){    int T;    //freopen("d:\\Test.txt","r",stdin);    cin>>T;    while(T--)    {        cin>>n>>s;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        sum[0]=a[0];        for(int i=1;i<n;i++)        {            sum[i]=sum[i-1]+a[i];        }        if(solve()==n+1) cout<<"0"<<endl;        else cout<<solve()<<endl;    }    return 0;}

Ruler Acquisition Method

(1) set two pointers, S and T, pointing to the first element of the series at the beginning, and sum = 0, Res = 0;

(2) As long as sum <s, an element is added to sum, and T is added to 1;

(3) until sum> = s, update res = min (Res, t-s );

(4) subtract an element from Sum, add s to 1, and execute (2 ).

The above process repeatedly advances the beginning and end of the interval to obtain the minimum interval that meets the conditions.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,m;int a[100005];void solve(){    int res=n+1;    int s=0,t=0,sum=0;    while(true)    {        while(t<n&&sum<m)        {            sum+=a[t++];        }        if(sum<m) break;        res=min(res,t-s);        sum-=a[s++];    }    if(res>n) res=0;    cout<<res<<endl;}int main(){int T;    //freopen("d:\\Test.txt","r",stdin);    cin>>T;    while(T--)    {        cin>>n>>m;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        solve();    }    return 0;}

Poj3061 subsequence (Binary prefix and method + ruler acquisition)