POJ 3320 Jessica ' s Reading problem (ruler)

Topic Links:

http://poj.org/problem?id=3320

Test instructions

A book has P page, each page has a knowledge point AI, there are different two pages on the same knowledge points,

Ask for a minimum number of consecutive pages of the book, can be all the knowledge points covered.

Ruler extraction:

The method of finding the minimum interval for satisfying a condition is called a ruler by repeatedly pushing the beginning and ending of the interval.

The degree of complexity of the ruler is O (n).

Initialize, start, end, sum = 0;

Then push end until the condition exits, if the interval has reached the end and is still not satisfied, exit

Then update the answer ans = min (ans,end-start);

Then start advances one, to deal with the effect of this time on sum.

The code is as follows:

#include <iostream> #include <cstring> #include <cstdio> #include <map>using namespace std; const int MAXN = 1e6+10;int A[maxn];int main () {    int n;    while (~SCANF ("%d", &n)) {        map<int, int >mp;        int num = 0;        for (int i=0;i<n;i++) {            scanf ("%d", a+i);            if (!mp[a[i]]) num++;            mp[a[i]]++;        }        int st=0,t=0,sum=0,ans = n;        Mp.clear ();        while (1) {            while (t<n&&sum<num) {                if (!mp[a[t++]]++)                    sum++;            }            if (sum<num) break;            ans = min (ans,t-st);            if (--mp[a[st++]]==0)                sum--;        }        printf ("%d\n", ans);    }    return 0;}

and a similar subject, POJ 3061.

Subsequence

Test instructions: Given a sequence, the minimum value of the length of successive sub-sequences with a sum not less than S is calculated

There are two ways to do this problem.

Method One:

Pretreatment, sum[i]=a[0]+....+a[i-1];

Then, in the interval [i,n] dichotomy, look for the subscript that is greater than or equal to the minimum value of Sum[i]+s,

Ans=min (ans,pos-i);

The code is as follows:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int MAXN = 1e5+10;int A[maxn],sum[maxn];int main () {    int t,s,n;    scanf ("%d", &t);    while (t--) {        scanf ("%d%d", &n,&s);        sum[0]=0;        for (int i=0;i<n;i++) {            scanf ("%d", a+i);            Sum[i+1]=sum[i]+a[i];        }        if (sum[n]<s) {            printf ("0\n");            Continue;        }        int ans = n;        for (int i=0;sum[i]+s<=sum[n];i++) {            int pos=lower_bound (sum+i,sum+n,sum[i]+s)-sum;            ans = min (ans,pos-i);        }        printf ("%d\n", ans);    }    return 0;}

Method Two,

Ruler extraction

The code is as follows:

#include <iostream> #include <cstring> #include <cstdio>using namespace std;const int maxn = 1e5+10;int A[maxn];int Main () {    int t,n,s;    scanf ("%d", &t);    while (t--) {        scanf ("%d%d", &n,&s);        for (int i=0;i<n;i++)            scanf ("%d", &a[i]);        int st=0,t=0,sum=0,ans= n+1;        while (1) {            while (t<n&&sum<s)                sum+=a[t++];            if (sum<s) break;            Ans=min (ans,t-st);            sum-=a[st++];        }        if (ans>n) {            printf ("0\n");            Continue;        }        else            printf ("%d\n", ans);    }    return 0;}

POJ 3320 Jessica ' s Reading problem (ruler)