HDU 4352 数位dp

题目链接

题意

分析

111111111

Code

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
using namespace std;
typedef long long ll;

int T, cases = 0, K, digit[20];
ll L, R, dp[20][1100][10];

ll dfs(int pos, int state, int len, int first, int limit)
{ // len:已有长度， state:状压，first:前导0
if (pos == -1) return len == K;
if (len > K) return 0;
if (!limit && dp[pos][state][K] != -1) return dp[pos][state][K];
int last = limit ? digit[pos] : 9;
ll ret = 0;
for (int i = 0; i <= last; ++i) {
int next_state = state, next_len = len;
if((i != 0 || (!first && i == 0)) && state < (1 << i)) { // i是最大数字
next_state += (1 << i);
next_len++;
} else if (((state >> i) & 1) == 0){
int k = i + 1;
while (k != 10 && ((state >> k) & 1) == 0) ++k; // 使子序列潜力变大
if (k != 10) {
next_state = next_state - (1 << k) + (1 << i);
}
}
ret += dfs(pos - 1, next_state, next_len, first && (i == 0), limit && (i == last));
}
if (!limit) dp[pos][state][K] = ret;
return ret;
}

ll solve(ll x)
{
if (x == 0) return 0;
memset(digit, 0, sizeof (digit));
int len = 0;
while (x) {
digit[len++] = x % 10;
x /= 10;
}
return dfs(len - 1, 0, 0, 1, 1);
}

int main()
{
memset(dp, -1, sizeof (dp));
scanf("%d", &T);
while (T--) {
scanf("%lld%lld%d", &L, &R, &K);
printf("Case #%d: %lld\n", ++cases, solve(R) - solve(L - 1));
}
return 0;
}