hdu 4669 Mutiples on a circle

来源：互联网发布：四大名助知乎编辑：程序博客网时间：2024/05/19 16:03

dp递推，通过一个状态往后加数不断递推出所有状态，并且不断取余

#include <cstdio>#include <cstring>#include <cmath>#define M 50005#define K 205int mod[M], sum[M], dp[M][K], p[200005];int n, k;void POW(){    p[0] = 1;    for(int i = 1; i <= 4*n; ++i) p[i] = (p[i-1]*10)%k;}int main (){    //freopen("in.txt","r",stdin);    //freopen("ou.txt","w",stdout);    while(scanf("%d%d",&n, &k)!=EOF)    {        POW();        for(int i=0; i<=n; ++i)            for(int j=0; j<=k; ++j) dp[i][j]=0;        for(int i = 1; i <= n; ++i)        {            scanf("%d",&sum[i]);            mod[i] =1+ log10(sum[i]*1.0);        }        int cnt=mod[1],s=sum[1]%k;        ++dp[1][s];        for(int i = 2; i <= n; ++i)        {            s = (sum[n-i+2]*p[cnt]+s)%k;            ++dp[1][s];            cnt+=mod[n-i+2];        }        int ans = dp[1][0];        for(int i = 2; i <= n; ++i)        {            ++dp[i][sum[i]%k];            for(int j = 0; j < k; ++j)                if(dp[i-1][j])                {                    int t = (j*p[mod[i]]+sum[i])%k;                    dp[i][t] += dp[i-1][j];                }            s=(s*p[mod[i]]+sum[i])%k;            dp[i][s]-=1;            ans+=dp[i][0];            s=(s-(sum[i]*p[cnt])%k+k)%k;        }        printf("%d\n",ans);    }    return 0;}