P2602 [ZJOI2010]数字计数

P2602 [ZJOI2010]数字计数

给定两个正整数a和b，求在[a,b]中的所有整数中，每个数码(digit)各出现了多少次。

Solution

一眼识数位dp
以 \(dp[i][j]\) 对某个指定的数码， 填了 \(i\) 位， 其中有 \(j\) 位填了特定数码的 总数码数
然后分一下前导零， 最高位， 数位dp即可
看代码

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
#define REP(i, x, y) for(LL i = (x);i <= (y);i++)
using namespace std;
LL RD(){
    LL out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const LL maxn = 19;
LL num[maxn];
LL mem[maxn];
LL dp[maxn][maxn];//填完i位，填了j个给定数码的数码总数
LL a, b;
LL DP(LL Index, LL sum, LL state, bool zero, bool limit){
	if(Index == 0)return sum;//填完一组新的，总数累加sum
	if(!zero && !limit && dp[Index][sum] != -1)return dp[Index][sum];
	LL ans = 0, up = limit ? num[Index] : 9, add = 0;
	REP(i, 0, up){
		if(state != 0 || (state == 0 && !zero))add = (i == state);
		ans += DP(Index - 1, sum + add, state, zero && i == 0, limit && i == num[Index]);
		}
	if(!zero && !limit)dp[Index][sum] = ans;
	return ans;
	}
LL get_num(LL x, LL state){
	LL len = 0;
	while(x){
		num[++len] = x % 10;
		x /= 10;
		}
	return DP(len, 0, state, 1, 1);
	}
void init(){
	a = RD(), b = RD();
	memset(dp, -1, sizeof(dp));
	}
void solve(){
	REP(i, 0, 9){
		printf("%lld ", get_num(b, i) - get_num(a - 1, i));
		}
	puts("");
	}
int main(){
	init();
	solve();
	return 0;
	}