Shovel Sale CodeForces - 899D (数位dp)

大意: n把铲子, 价格1,2,3,...n, 求有多少个二元组(x,y), 满足x+y末尾数字9的个数最多.

枚举最高位, 转化为从[1,n]中选出多少个二元组和为$x$, 枚举较小的数

若$n\ge \lfloor\frac{x}{2}\rfloor$答案为$\lfloor\frac{x}{2}\rfloor$, 否则为$n-\lfloor\frac{x}{2}\rfloor$

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 100;
ll n, tot, f[N], num[N];
int id(int x) {
	return num[upper_bound(f+1,f+1+9,x)-f-1];
}

int main() {
	f[1] = 5;
	REP(i,2,9) f[i]=f[i-1]*10;
	num[1] = 9;
	REP(i,2,9) num[i]=num[i-1]*10+9;
	cin>>n, tot = id(n);
	if (tot==0) return cout<<n*(n-1)/2<<endl,0;
	ll ans = 0;
	REP(i,0,9) {
		ll num = (ll)i*(tot+1)+tot;
		if (n<=num/2) break;
		else if (n>=num) ans+=num/2;
		else ans+=n-num/2;
	}
	printf("%lld\n",ans);
}