HDU 3555 Bomb 数位dp

题目：

题意：

思路：

//300+ms
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int N = 50 + 10, INF = 0x3f3f3f3f;

ll dp
[10][2]; //dp[i][j][k]：在无限制取值范围条件下第i位后前驱为j时状态为k时的方案数
int dig
;//分离后的数字
ll dfs(int pos, int pre, int status, int limit)
{//pos是当前处理的位置，pre是前一个处理的位上的数字，status是有没有已经出现了49这个子串，limit表示对于当前位数字的取值范围有没有限制
if(pos < 0) return status;
if(! limit && dp[pos][pre][status] != -1) return dp[pos][pre][status];
int en = limit ? dig[pos] : 9;
ll ans = 0;
for(int i = 0; i <= en; i++)
ans += dfs(pos-1, i, status || (pre == 4 && i == 9), limit && i == en);
if(! limit) dp[pos][pre][status] = ans;
return ans;
}
int main()
{
int t;
ll n;
scanf("%d", &t);
while(t--)
{
scanf("%lld", &n);
int tot = 0;
while(n) dig[tot++] = n % 10, n /= 10;
memset(dp, -1, sizeof dp);
ll ans = dfs(tot-1, 0, 0, 1);
printf("%lld\n", ans);
}
return 0;
}

//70ms
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int N = 30 + 10, INF = 0x3f3f3f3f;

ll dp
[2];
ll p
;
int dig
;
ll n;

void table()
{
p[0] = 1;
for(int i = 1; i < N; i++) p[i] = p[i-1] * 10;
}
ll dfs(int pos, bool is4, bool limit)
{
if(pos < 0) return 0;
if(! limit && dp[pos][is4] != -1) return dp[pos][is4];
int en = limit ? dig[pos] : 9;
ll ans = 0;
for(int i = 0; i <= en; i++)
if(is4 && i == 9) ans += limit ? n % p[pos] + 1 : p[pos];//直接计算无需递归，效率提升很多
else ans += dfs(pos-1, i == 4, limit && i == en);
if(! limit) dp[pos][is4] = ans;
return ans;
}
int main()
{
table();
int t;
scanf("%d", &t);
while(t--)
{
scanf("%lld", &n);
ll m = n;
int k = 0;
while(m) dig[k++] = m % 10, m /= 10;
memset(dp, -1, sizeof dp);
ll ans = dfs(k-1, 0, 1);
printf("%lld\n", ans);
}
return 0;
}

//这个是求不含49的个数，然后用总数减去
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int N = 30 + 10, INF = 0x3f3f3f3f;

ll dp
[2];
ll p
;
int dig
;
ll n;

void table()
{
p[0] = 1;
for(int i = 1; i < N; i++) p[i] = p[i-1] * 10;
}
ll dfs(int pos, bool is4, bool limit)
{
if(pos < 0) return 1;
if(! limit && dp[pos][is4] != -1) return dp[pos][is4];
int en = limit ? dig[pos] : 9;
ll ans = 0;
for(int i = 0; i <= en; i++)
if(is4 && i == 9) ans += 0;
else ans += dfs(pos-1, i == 4, limit && i == en);
if(! limit) dp[pos][is4] = ans;
return ans;
}
int main()
{
table();
int t;
scanf("%d", &t);
while(t--)
{
scanf("%lld", &n);
ll m = n;
int k = 0;
while(m) dig[k++] = m % 10, m /= 10;
memset(dp, -1, sizeof dp);
ll ans = dfs(k-1, 0, 1);
printf("%lld\n", n - ans + 1);
}
return 0;
}