题目描述:
- 有nnn种物品,第i种物品有aia_iai个,不同种类物品可以相互区分但相同种类的物品无法区分。从这些物品中取出mmm个的话,有多少种取法,求出方案数模MMM的余数。
限制条件:
- 1≤m,n,ai≤10001≤m,n,a_i≤10001≤m,n,ai≤1000
- 0≤M≤100000≤M≤100000≤M≤10000
样例说明:
- 输入:
- 输出:
- 6=(0+0+3=0+1+2=0+2+1=1+0+2=1+1+1=1+2+0=m=3)
- 注意:0+1+2表示:第1/2类物品取0个,第三类物品取3个。根据这个含义0+1+2和0+2+1不是重复的。
动态规划:
- 步骤:
- dp数组含义:dp[i][j]dp[i][j]dp[i][j]=从前iii种物品中取出jjj个的组合数。取不出令为0.
- 初始条件:
- dp[0−n][0]=1dp[0-n][0]=1dp[0−n][0]=1
- dp[1][1−m]={1a[1]≥j0a[1]<jdp[1][1-m]=\begin{cases} 1 & a[1]≥j \\ 0 & a[1]<j \end{cases}dp[1][1−m]={10a[1]≥ja[1]<j
- 递推公式:dp[i][j]=∑k=0min(j,a[i])dp[i−1][j−k]dp[i][j]=∑_{k=0}^{min(j,a[i])}dp[i-1][j-k]dp[i][j]=k=0∑min(j,a[i])dp[i−1][j−k]
- 递推方向:从上向下
- 时间复杂度 :O(nm2)O(nm^2)O(nm2)
- 优化 :
- 根据一般的经验,在只使用从上到下的递推的动态规划中如果能引入左到右的辅助递推,时间复杂度为更低。根据这一思想,我们化解上面的递推公式:∑k=0min(j,a[i])dp[i−1][j−k]=∑k=1min(j,a[i])dp[i−1][j−k]+dp[i−1][j]∑_{k=0}^{min(j,a[i])}dp[i-1][j-k]=∑_{k=1}^{min(j,a[i])}dp[i-1][j-k]+dp[i-1][j]k=0∑min(j,a[i])dp[i−1][j−k]=k=1∑min(j,a[i])dp[i−1][j−k]+dp[i−1][j]=∑k=0min(j−1,a[i]−1)dp[i−1][j−(1+k)]+dp[i−1][j]=∑_{k=0}^{min(j-1,a[i]-1)}dp[i-1][j-(1+k)]+dp[i-1][j]=k=0∑min(j−1,a[i]−1)dp[i−1][j−(1+k)]+dp[i−1][j]由于dp[i][j−1]=∑k=0min(j−1,a[i])dp[i−1][j−1−k]=∑k=0min(j−1,a[i])dp[i−1][j−(1−k)]dp[i][j-1]=∑_{k=0}^{min(j-1,a[i])}dp[i-1][j-1-k]=∑_{k=0}^{min(j-1,a[i])}dp[i-1][j-(1-k)]dp[i][j−1]=k=0∑min(j−1,a[i])dp[i−1][j−1−k]=k=0∑min(j−1,a[i])dp[i−1][j−(1−k)]当j≤a[i]j≤a[i]j≤a[i]时则有j−1≤a[i]−1j-1≤a[i]-1j−1≤a[i]−1和j−1≤a[i]j-1≤a[i]j−1≤a[i],所以min(j−1,a[i]−1)=min(j−1,a[i])=j−1min(j-1,a[i]-1)=min(j-1,a[i])=j-1min(j−1,a[i]−1)=min(j−1,a[i])=j−1 :∑k=0min(j−1,a[i]−1)dp[i−1][j−(1+k)]=∑k=0min(j−1,a[i])dp[i−1][j−(1+k)]=dp[i][j−1]∑_{k=0}^{min(j-1,a[i]-1)}dp[i-1][j-(1+k)]=∑_{k=0}^{min(j-1,a[i])}dp[i-1][j-(1+k)]=dp[i][j-1]k=0∑min(j−1,a[i]−1)dp[i−1][j−(1+k)]=k=0∑min(j−1,a[i])dp[i−1][j−(1+k)]=dp[i][j−1]当j>a[i]j>a[i]j>a[i],所以j−1>a[i]−1j-1>a[i]-1j−1>a[i]−1和j−1≥a[i]j-1≥a[i]j−1≥a[i],故min(j−1,a[i]−1)=a[i]−1,min(j−1,a[i])=a[i]min(j-1,a[i]-1)=a[i]-1,min(j-1,a[i])=a[i]min(j−1,a[i]−1)=a[i]−1,min(j−1,a[i])=a[i]则:∑k=0min(j−1,a[i]−1)dp[i−1][j−(1+k)]=∑k=0min(j−1,a[i])dp[i−1][j−(1+k)]−dp[i−1][j−1−a[i])]∑_{k=0}^{min(j-1,a[i]-1)}dp[i-1][j-(1+k)]=∑_{k=0}^{min(j-1,a[i])}dp[i-1][j-(1+k)]-dp[i-1][j-1-a[i])]k=0∑min(j−1,a[i]−1)dp[i−1][j−(1+k)]=k=0∑min(j−1,a[i])dp[i−1][j−(1+k)]−dp[i−1][j−1−a[i])]=dp[i][j−1]−dp[i−1][j−1−a[i])]=dp[i][j-1]-dp[i-1][j-1-a[i])]=dp[i][j−1]−dp[i−1][j−1−a[i])]
- 时间复杂度 :O(nm)O(nm)O(nm)
- 结果:dp[n][m]dp[n][m]dp[n][m]
代码:
#include <iostream>
#define Max_N 1005
using namespace std;
int n,m,M;
int a[Max_N];
int dp[Max_N][Max_N];
void solve()
{
for(int i=0; i<=n; i++)
dp[i][0]=1;
for(int j=1;j<=m;j++)
dp[0][j]=0;
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
{
dp[i][j]=dp[i][j-1]+dp[i-1][j];
if (j>a[i])
dp[i][j]-=dp[i-1][j-1-a[i]];
dp[i][j]=dp[i][j]%M;
}
cout<<dp[n][m]<<endl;
}
int main()
{
cin>>n>>m;
for(int i=1;i<=n;i++)
cin>>a[i];
cin>>M;
solve();
return 0;
}