BZOJ原题链接
洛谷原题链接
又是套记搜模板的时候。。
对(0sim 9)单独统计。
定义(f[pos][sum])，即枚举到第(pos)位，前面枚举的所有位上是当前要统计的数的个数之和为(sum)。
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int N = 13;
ll f[N][N];
int a[N], nw;
inline ll re()
{
ll x = 0;
char c = getchar();
bool p = 0;
for (; c < '0' || c > '9'; c = getchar())
p |= c == '-';
for (; c >= '0' && c <= '9'; c = getchar())
x = x * 10 + c - '0';
return p ? -x : x;
}
ll dfs(int pos, int s, int lm, int zero)
{
if (pos < 0)
return s;
if (!lm && !zero && f[pos][s] > -1)
return f[pos][s];
int i, k = lm ? a[pos] : 9;
ll S = 0;
for (i = 0; i <= k; i++)
S += dfs(pos - 1, s + ((!nw && !i && !zero) || (nw && i == nw)), lm && i == a[pos], zero && !i);
if (!lm && !zero)
return f[pos][s] = S;
return S;
}
ll calc(ll x)
{
int k = 0;
memset(f, -1, sizeof(f));
while (x > 0)
{
a[k++] = x % 10;
x /= 10;
}
return dfs(k - 1, 0, 1, 1);
}
int main()
{
ll x, y;
x = re();
y = re();
for (nw = 0; nw <= 9; nw++)
printf("%lld ", calc(y) - calc(x - 1));
return 0;
}