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  • P2602 [ZJOI2010]数字计数

    https://www.luogu.org/problemnew/show/P2602

    数位dp

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 #define ll long long
     4 const double minv=1e-10;
     5 
     6 ll g[10],shi[15];
     7 
     8 void cal(ll a,ll c,int ori)
     9 {
    10     int i,w,s;
    11     w=(log(a+minv)/log(10));
    12     i=w;
    13     while (i>=0)
    14         g[0]-=shi[i--]*c;
    15     while (w>=0)
    16     {
    17         s=a/shi[w];
    18         if (w!=0)
    19             for (i=0;i<10;i++)
    20                 g[i]+=s*shi[w-1]*w*c;
    21         for (i=0;i<s;i++)
    22             g[i]+=shi[w]*c;
    23         g[s]+=(a%shi[w]+1)*c;
    24         a=a%shi[w];
    25         w--;
    26     }
    27 }
    28 
    29 int main()
    30 {
    31     ll a,b;
    32     int i;
    33     scanf("%lld%lld",&a,&b);
    34     shi[0]=1;
    35     for (i=1;i<=12;i++)
    36         shi[i]=shi[i-1]*10;
    37     cal(b,1,0);
    38     cal(a-1,-1,0);
    39     for (i=0;i<10;i++)
    40     {
    41         printf("%lld",g[i]);
    42         if (i!=10)
    43             printf(" ");
    44     }
    45     return 0;
    46 }
    47 /*
    48 1 999
    49 5 10
    50 10 20
    51 10 19
    52 1 1000000000000
    53 99 1000000000000
    54 */
    55 /*
    56 13 99
    57 8 15 18 19 19 19 19 19 19 19
    58 13 550
    59 104 210 213 214 214 156 104 104 104 104
    60 2 110
    61 21 32 21 21 21 21 21 21 21 21
    62 1 100000000
    63 68888897 80000001 80000000 80000000 80000000 80000000 80000000 80000000 80000000 80000000
    64 */

    验证

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 #define ll long long
     4 #define minv 1e-6
     5 #define inf 1e9
     6 #define pi 3.1415926536
     7 #define nl 2.7182818284
     8 const ll mod=1e9+7;//998244353
     9 const int maxn=1e5+10;
    10 
    11 ll g[10];
    12 
    13 int main()
    14 {
    15     int i,j;
    16     ll a,b;
    17     scanf("%lld%lld",&a,&b);
    18     for (i=a;i<=b;i++)
    19     {
    20         j=i;
    21         while (j)
    22         {
    23             g[j%10]++;
    24             j/=10;
    25         }
    26     }
    27     for (i=0;i<10;i++)
    28         printf("%lld ",g[i]);
    29     return 0;
    30 }
    31 /*
    32 13 99
    33 8 15 18 19 19 19 19 19 19 19
    34 13 550
    35 104 210 213 214 214 156 104 104 104 104
    36 2 110
    37 21 32 21 21 21 21 21 21 21 21
    38 1 100000000
    39 68888897 80000001 80000000 80000000 80000000 80000000 80000000 80000000 80000000 80000000
    40 */
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  • 原文地址:https://www.cnblogs.com/cmyg/p/9882743.html
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