限制条件的DP总结

　　对于什么至少K个连续正面朝上求排列种数的题目，这样的限制条件DP题目，可以考虑转化为至多 V 个连续朝上的，这样答案，只需 solve(N) - solve(k-1) 即可得到。

然后对于状态的转移 ：一般可以类似这样  int  dp[MAXS][2];  dp[i][0] :表示第i个位置正面朝上，dp[i][1]:表示反面朝上的方法数(且最多有k张连续的正面的种数)

具体下面有两道例题，思想很相似。

Uva 10328  (这道题好需要大数才行，数据很大，需要个大数类模板 ，推荐大佬的一个模板： https://blog.csdn.net/code4101/article/details/23020525)

https://blog.csdn.net/cc_again/article/details/24844911

//#include <bits/stdc++.h>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>  
#include <string>  
#include <cstring>  
#include <cstdio>  
using namespace std;  
  
const int maxn = 1000;  
const int MAXS = 106;
  
struct bign{  
    int d[maxn], len;  
  
    void clean() { while(len > 1 && !d[len-1]) len--; }  
  
    bign()          { memset(d, 0, sizeof(d)); len = 1; }  
    bign(int num)   { *this = num; }   
    bign(char* num) { *this = num; }  
    bign operator = (const char* num){  
        memset(d, 0, sizeof(d)); len = strlen(num);  
        for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';  
        clean();  
        return *this;  
    }  
    bign operator = (int num){  
        char s[20]; sprintf(s, "%d", num);  
        *this = s;  
        return *this;  
    }  
  
    bign operator + (const bign& b){  
        bign c = *this; int i;  
        for (i = 0; i < b.len; i++){  
            c.d[i] += b.d[i];  
            if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;  
        }  
        while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;  
        c.len = max(len, b.len);  
        if (c.d[i] && c.len <= i) c.len = i+1;  
        return c;  
    }  
    bign operator - (const bign& b){  
        bign c = *this; int i;  
        for (i = 0; i < b.len; i++){  
            c.d[i] -= b.d[i];  
            if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;  
        }  
        while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;  
        c.clean();  
        return c;  
    }  
    bign operator * (const bign& b)const{  
        int i, j; bign c; c.len = len + b.len;   
        for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)   
            c.d[i+j] += d[i] * b.d[j];  
        for(i = 0; i < c.len-1; i++)  
            c.d[i+1] += c.d[i]/10, c.d[i] %= 10;  
        c.clean();  
        return c;  
    }  
    bign operator / (const bign& b){  
        int i, j;  
        bign c = *this, a = 0;  
        for (i = len - 1; i >= 0; i--)  
        {  
            a = a*10 + d[i];  
            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;  
            c.d[i] = j;  
            a = a - b*j;  
        }  
        c.clean();  
        return c;  
    }  
    bign operator % (const bign& b){  
        int i, j;  
        bign a = 0;  
        for (i = len - 1; i >= 0; i--)  
        {  
            a = a*10 + d[i];  
            for (j = 0; j < 10; j++) if (a < b*(j+1)) break;  
            a = a - b*j;  
        }  
        return a;  
    }  
    bign operator += (const bign& b){  
        *this = *this + b;  
        return *this;  
    }  
  
    bool operator <(const bign& b) const{  
        if(len != b.len) return len < b.len;  
        for(int i = len-1; i >= 0; i--)  
            if(d[i] != b.d[i]) return d[i] < b.d[i];  
        return false;  
    }  
    bool operator >(const bign& b) const{return b < *this;}  
    bool operator<=(const bign& b) const{return !(b < *this);}  
    bool operator>=(const bign& b) const{return !(*this < b);}  
    bool operator!=(const bign& b) const{return b < *this || *this < b;}  
    bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}  
  
    string str() const{  
        char s[maxn]={};  
        for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';  
        return s;  
    }  
};  
  
istream& operator >> (istream& in, bign& x)  
{  
    string s;  
    in >> s;  
    x = s.c_str();  
    return in;  
}  
  
ostream& operator << (ostream& out, const bign& x)  
{  
    out << x.str();  
    return out;  
}  

int n,k;
/*
  转化为至多，slove(v):表示只多有v个连续正面朝上硬币。
  那么答案 ans = slove (n) - slove(k-1);
 */
bign dp[MAXS][2];// dp[i][0] :表示第i个位置正面朝上，dp[i][1]:表示反面朝上的方法数(且最多有k张连续的正面的种数)

bign solve(int v)
{
   dp[0][0] = 0;
   dp[0][1] = 1;
   for (int i = 1; i <= n; ++i)
   {
        /* code */
      bign sum = dp[i-1][0] + dp[i-1][1];
      dp[i][1] = sum;
      if (i <= v) dp[i][0] = sum;
      else if (i == v+1) dp[i][0] = sum -1;
      else  dp[i][0] = (sum - dp[i-v-1][1]); 
   }
   return dp[n][0] + dp[n][1];//之所以返回的是dp[n][0] + dp[n][1],因为最后方法总数                        
   //不就是n个位置放完吗，也就是为第n个位置放正or反这两种情况
}

int main(int argc, char const *argv[])
{
    
    while(~scanf ("%d%d",&n,&k))
    {
      bign ans = solve(n);
      cout << (ans - solve(k-1))<<endl;
    }
    return 0;
}

ZOJ 3747

https://blog.csdn.net/cc_again/article/details/24841249

#include <bits/stdc++.h>
#define MAX 1000000+100
#define MOD 1000000007
using namespace std;
/*
zoj 3747  dp 递推 
条件限制DP  将条件都转化为至多。 
 
*/
typedef long long ll;
int n,m,k; 
ll dp[MAX][3];//dp[i][0]:表示第 i 个位置 放 0 （G士兵） 的方法数目，一直放满 N 个位置 


ll solve(int u,int v)
{
  //初始化
  dp[0][0] = dp[0][1] = 0;
  dp[0][2] = 1;
  ll sum = 0;
  for (int i=1;i<=n;++i)
   {
        sum = ( dp[i-1][0] + dp[i-1][1] + dp[i-1][2] )%MOD;
        dp[i][2] = sum;
        // 对于G士兵 
        if ( i <= u)// 此时连续G个数没有超过可以随便放，所以种数 =sum 
          dp[i][0] = sum;
        else if ( i == u +1) dp[i][0] = sum-1;//此时刚好超一个，那么我们就减去 
        else dp[i][0] = (sum - dp[i-u-1][1] - dp[i-u-1][2] ) % MOD;
        //对于R士兵
      if ( i <= v)
          dp[i][1] = sum;
        else if ( i == v +1) dp[i][1] = sum-1;
        else dp[i][1] = ( sum - dp[i-v-1][0] - dp[i-v-1][2] ) % MOD;
   } 
   return  ( ( dp[n][0] + dp[n][1] + dp[n][2] ) % MOD ); 
}


int main ()
{
  while(~scanf("%d%d%d",&n,&m,&k))
  {
    ll ans = solve(n,k);
    cout << ( ( (ans - solve(m-1,k))%MOD +MOD ) % MOD) << endl;
  }
  
  return 0;
}