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题目链接:https://ac.nowcoder.com/acm/contest/338/F
题意:就是让你统计1~n中包含520的数的个数,n<= 1e18
思路:数位DP中比较简单的了,dp[pos][pre1][pre2]代表着第pos位,前缀是pre1和pre2的不包含520的个数,最后用总数减去包含520的个数就是答案。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
#define ls rt << 1
#define rs rt << 1|1
#define mid ((l + r) >> 1)
#define lson l, mid, ls
#define rson mid + 1, r, rs
const int maxn = 20;
const int mod = 1e9 + 7;
ll dp[maxn][12][12];
int a[maxn];
ll dfs(int pos, int pre1, int pre2, int lead, int limit)
{
if(pos == 0) return 1;
if(!limit && ! lead && dp[pos][pre1][pre2] != -1)
return dp[pos][pre1][pre2];
int up = limit ? a[pos] : 9;
ll ret = 0;
for(int i = 0; i <= up; ++i)
{
if(pre1 == 5 && pre2 == 2 && i == 0) continue;
ret += dfs(pos - 1, pre2, i, lead && (i == 0), limit && (i == up));
}
if(!limit && !lead)
dp[pos][pre1][pre2] = ret;
return ret;
}
ll slove(ll x)
{
int tot = 0;
while(x)
{
a[++tot] = x % 10;
x /= 10;
}
memset(dp, -1, sizeof(dp));
return dfs(tot, 0, 0, 1, 1);
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
ll n;
scanf("%lld", &n);
ll ans = n - slove(n) + 1;
printf("%lld\n", ans);
}
return 0;
}