DP - Counting 计数


http://luoshaochuan.com/2017/07/14/动态规划总结(五)/

划分数

有N个无区别的物品,将它们划为不超过M堆,求划分方案数。

分析

dp[i][j]表示j个物品划为i+1堆,a1,a2,…,ai,则若存在ai=0,则共有dp[i-1][j]堆,若任意ai!=0,共有dp[i][j-i]堆。所以:
1
dp[i][j]=dp[i-1][j]+dp[i][j-1]

2.多重集合数

有N种物品,每种物品有Ai个,从这些物品中取出M个,求方案数。

分析

dp[i+1][j]表示前i种物品中取出j个的方案数。则对于第i种物品,可以取出0,1,…Ai个。所以
1
dp[i+1][j]=dp[i][j]+dp[i][j-1]+...+dp[i][j-Ai]=\sum_{k=0}^{min(j,Ai)}dp[i][j-k]

优化

这种求和的动态规划,通常可以把递推式进行变形,从而降低复杂度。
1
\sum_{k=0}^{min(j,Ai)}dp[i]=\sum_{k=0}^{min(j-1,Ai)}dp[i][(j-1)-k]+dp[i][j]-dp[i][j-1-Ai]=dp[i+1][j-1]+dp[i][j]-dp[i]



https://blog.csdn.net/kangyan__/article/details/66969328
描述:计算从1到n中,每个数字(0到9)出现的次数
其中sum[j]和dp[i]表示:数字i中 j 的个数;比如sum[1]和dp[5]就可以表示:5中1的个数为0;sum[1]和dp[11]就可以表示:11中1的个数为2;

int dp[100001];
int main()
{
    int n,sum[10];
    while(~scanf("%d",&n))
    {
        memset(sum,0,sizeof(sum));
        int j=0;
        while(j<10)
        {
            memset(dp,0,sizeof(dp));
            dp[j] = 1;
            for (int i = 1; i <= n; i++)
            {
                dp[i] = dp[i%10];
                if(i/10)
                    dp[i] += dp[i/10];
                sum[j] += dp[i];
            }
            j++;
        }
        for(int i=0;i<10;i++)
            printf("%d:%d\n",i,sum[i]);
    }
    return 0;
}


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