尺取法 poj 2566

尺取法：顾名思义就是像尺子一样一段一段去取，保存每次的选取区间的左右端点。然后一直推进

解决问题的思路：

先移动右端点，右端点推进的时候一般是加
然后推进左端点，左端点一般是减

poj 2566

题意：从数列中找出连续序列，使得和的绝对值与目标数之差最小

思路：

在原来的数列开头添加一个0
每次找到的区间为 [min(i,j)+1,max(i,j)]

应用尺取法的代码：

 while (r <=n)
                       {
                              int sub = pre[r].sum - pre[l].sum;
                              while (abs(sub - t) < Min)
                              {
                                      Min = abs(sub - t);
                                      ansl= min(pre[l].id, pre[r].id) + 1;
                                      ansr = max(pre[l].id, pre[r].id);
                                      ans = sub;
                              }
                              if (sub < t)
                                      r++;
                              else if (sub > t) l++;
                              else break;
                              if (l == r) r++;
                       }

解决问题的代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x7fffffff;
int n, k;
int a[N];
struct node {
        int sum, id;
}pre[N];
bool cmp(node a, node b)
{
        return a.sum < b.sum;
}
int main()
{
        while (scanf("%d%d", &n, &k) != EOF)
        {
               if (n == 0 && k == 0) break;
               pre[0].id = 0, pre[0].sum = 0;
               for (int i = 1; i <= n; i++)
               {
                       scanf("%d", &a[i]);
                       pre[i].id = i;
                       pre[i].sum = pre[i - 1].sum + a[i];
               }
               sort(pre, pre + n + 1, cmp);
               for (int i = 0; i < k; i++)
               {
                       int t;
                       scanf("%d", &t);
                       int Min = INF;
                       int l = 0, r = 1;
                       int ans, ansl, ansr;
                       while (r <=n)
                       {
                              int sub = pre[r].sum - pre[l].sum;
                              while (abs(sub - t) < Min)
                              {
                                      Min = abs(sub - t);
                                      ansl= min(pre[l].id, pre[r].id) + 1;
                                      ansr = max(pre[l].id, pre[r].id);
                                      ans = sub;
                              }
                              if (sub < t)
                                      r++;
                              else if (sub > t) l++;
                              else break;
                              if (l == r) r++;
                       }
                       printf("%d %d %d\n", ans, ansl, ansr);
               }
        }
        return 0;
}

poj 2739

题意：将一个整数分解为连续的素数之和，有多少种分法?

思路：

打表，先打出素数表
然后依次查询是否满足条件

用尺取法的代码：

 for (;;) {
                       while (r < size&&sum < n)
                       {
                              sum += primes[r++];
                       }
                       if (sum < n) break;
                       else if (sum == n) result++;
                       sum -= primes[l++];
               }

解决问题的代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
#define maxn 10000+16
vector<int> primes;
vector<bool> is_prime;
void init_prime()
{
        is_prime = vector<bool>(maxn + 1, true);
        is_prime[0] = is_prime[1] = false;
        for (int i = 2; i < maxn; i++)
        {
               if (is_prime[i])
               {
                       primes.push_back(i);
                       for (int j = i * 2; j < maxn; j += i)
                       {
                              is_prime[j] = false;
                       }
               }
        }
}
int main()
{
        int n;
        init_prime();
        int size = primes.size();
        while (cin >> n && n)
        {
               int result = 0, sum = 0;
               int l = 0,r = 0;
               for (;;) {
                       while (r < size&&sum < n)
                       {
                              sum += primes[r++];
                       }
                       if (sum < n) break;
                       else if (sum == n) result++;
                       sum -= primes[l++];
               }
               printf("%d\n", result);
        }
        return 0;
}

poj 2100

题意：将一个整数分解为连续数平方之和，有多少种分法？

解决问题的思路：

右端点移动，sum+=r*r;
如果满足答案就 push (l,r)
左端点移动 sum-=l*l;

用尺取法的代码：

 for (;;)
        {
               while (sum < n)
               {
                       sq = r * r;
                       sum += sq;
                       r++;
               }
               if (sq > n) break;
               else if (sum == n) ans.push_back(make_pair(l, r));
               sum -= l * l;
               l++;
        }

解决问题的代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

typedef long long ll;
vector<pair<ll, ll>> ans;
void solve(ll n)
{
    ll l = 1, r = 1, sum = 0, sq;
    for (;;)
    {
        while (sum < n)
        {
            sq = r * r;
            sum += sq;
            r++;
        }
        if (sq > n) break;
        else if (sum == n) ans.push_back(make_pair(l, r));
        sum -= l * l;
        l++;
    }
    ll size = ans.size();
    printf("%lld\n", size);
    for (ll i = 0; i < size; i++)
    {
        ll ansr = ans[i].second;
        ll ansl = ans[i].first;
        printf("%lld",ansr-ansl);
        for (ll j=ansl;j<ansr;j++)
            printf(" %lld", j);
        printf("\n");
    }
}

int main()
{
    ll n;
    while (scanf("%lld", &n) != EOF)
    {
        if (n == 0) break;
        solve(n);
    }
    return 0;
}

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