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  • bzoj1833 [ZJOI2010]count 数字计数

    题目链接

    我是DP弱者!!!我是DP弱者!!!我是DP弱者!!!

    调了好久,数位DP好恼火QAQ

     1 #include<algorithm>
     2 #include<iostream>
     3 #include<cstdlib>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<string>
     7 #include<cmath>
     8 #include<ctime>
     9 #include<queue>
    10 #include<stack>
    11 #include<map>
    12 #include<set>
    13 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
    14 #define re(i,l,r) for(int i=(l);i<=(r);i++)
    15 #define Clear(a,b) memset(a,b,sizeof(a))
    16 #define inout(x) printf("%d",(x))
    17 #define douin(x) scanf("%lf",&x)
    18 #define strin(x) scanf("%s",(x))
    19 #define LLin(x) scanf("%lld",&x)
    20 #define op operator
    21 #define CSC main
    22 typedef unsigned long long ULL;
    23 typedef const int cint;
    24 typedef long long LL;
    25 using namespace std;
    26 void inin(int &ret)
    27 {
    28     ret=0;int f=0;char ch=getchar();
    29     while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();}
    30     while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar();
    31     ret=f?-ret:ret;
    32 }
    33 LL ten[13];
    34 LL dp[13][10][10];
    35 void getpre()
    36 {
    37     ten[0]=1;
    38     re(i,1,12)ten[i]=ten[i-1]*10;
    39     re(i,0,9)dp[1][i][i]=1;
    40     re(i,2,12)re(j,0,9)re(k,0,9)
    41     {
    42         dp[i][j][k]=(j==k)?ten[i-1]:0;
    43         re(t,0,9)dp[i][j][k]+=dp[i-1][t][k];
    44     }
    45 }
    46 int getbit(LL x)
    47 {
    48     int ret=0;
    49     while(x)
    50         x/=10,ret++;
    51     return ret;
    52 }
    53 LL ans[10];
    54 int num[15];
    55 void calc(LL temp,int bit,bool qian)
    56 {
    57     if(!bit)return;
    58     if(qian)
    59         re(i,0,num[bit]-1)
    60             re(j,0,9)
    61                 ans[j]+=dp[bit][i][j];
    62     else re(i,1,num[bit]-1)re(j,0,9)ans[j]+=dp[bit][i][j];
    63     ans[num[bit]]+=temp%ten[bit-1]+1;
    64     calc(temp,bit-1,1);
    65 }
    66 void calc2(LL temp,int bit,bool qian)
    67 {
    68     if(!bit)return;
    69     if(qian)
    70         re(i,0,num[bit]-1)
    71             re(j,0,9)
    72                 ans[j]-=dp[bit][i][j];
    73     else re(i,1,num[bit]-1)re(j,0,9)ans[j]-=dp[bit][i][j];
    74     ans[num[bit]]-=temp%ten[bit-1]+1;
    75     calc2(temp,bit-1,1);
    76 }
    77 int main()
    78 {
    79     getpre();
    80     LL l,r;
    81     scanf("%lld%lld",&l,&r);
    82     int len=0;
    83     LL temp=r;
    84     while(r)num[++len]=r%10,r/=10;
    85     calc(temp,len,0);
    86     re(i,1,len-1)re(j,0,9)re(k,1,9)ans[j]+=dp[i][k][j];
    87     len=0,temp=l-1;l--;
    88     while(l)num[++len]=l%10,l/=10;
    89     calc2(temp,len,0);
    90     re(i,1,len-1)re(j,0,9)re(k,1,9)ans[j]-=dp[i][k][j];
    91     re(i,0,9)printf("%lld%c",ans[i],i==9?'
    ':' ');
    92      return 0;
    93 }
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  • 原文地址:https://www.cnblogs.com/HugeGun/p/5267345.html
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