51 Nod 1042 数位dp

1042 数字0-9的数量

1. 1 秒
2. 131,072 KB
3. 10 分
4. 2 级题

 

给出一段区间a-b，统计这个区间内0-9出现的次数。

比如 10-19，1出现11次（10,11,12,13,14,15,16,17,18,19,其中11包括2个1)，其余数字各出现1次。

 

 

输入

两个数a,b（1 <= a <= b <= 10^18)

输出

输出共10行，分别是0-9出现的次数

输入样例

10 19

输出样例

1
11
1
1
1
1
1
1
1
1

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 20005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int len;
ll a, b;
int num[20];
ll dp[20][20002];

ll dfs(int pos, int limit, int lead, ll sum,int dig) {
	if (pos == 0)return sum;
	if (!limit&&lead&&dp[pos][sum] != -1)return dp[pos][sum];
	ll ans = 0;
	int up = limit ? num[pos] : 9;
	for (int i = 0; i <= up; i++) {
		ans += dfs(pos - 1, limit && (i == up), lead || i, sum + ((i == dig)&&(lead||i)), dig);
	}
	if (lead && !limit)dp[pos][sum] = ans;
	return ans;

}

ll sol(ll a, ll dig) {
	mclr(dp, -1); len = 0;
	while (a) {
		num[++len] = a % 10; a /= 10;
	}
	ll ans = 0;
	ans = dfs(len, 1, 0, 0, dig);
	return ans;
}
int main()
{
	//	ios::sync_with_stdio(0);
	rdllt(a); rdllt(b);
	for (int i = 0; i <= 9; i++) {
		printf("%lld\n", sol(b, i) - sol(a - 1, i));
	}
	return 0;
}