Sums of Digits CodeForces - 509C (贪心,模拟)

大意: 一个未知严格递增数组$a$, 给定每个数的数位和, 求$a[n]$最小的数组$a$

#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 = 310, M = 1e4+10;
int n, a[N];
int num[N][M+1];

void solve(int id, int s) {
	int sum = 0, pos = 0, res = 0;
	REP(i,1,M) { 
		if (num[id-1][i]!=9) {
			int l = num[id-1][i]+1+sum;
			int r = 9*(M-i+1)+sum;
			if (l<=s&&s<=r) { 
				pos = i, res = sum;
			}
		}
		sum += num[id-1][i];
	}
	REP(i,1,pos-1) num[id][i]=num[id-1][i];
	PER(i,num[id-1][pos]+1,9) {
		int l=i+res;
		int r=i+9*(M-pos)+res;
		if (l<=s&&s<=r) { 
			num[id][pos]=i;
		}
	}
	res += num[id][pos];
	REP(i,pos+1,M) {
		PER(j,0,9) {
			int l=res+j,r=j+9*(M-i)+res;
			if (l<=s&&s<=r) num[id][i]=j;
		}
		res += num[id][i];
	}
}


int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a+i);
	REP(i,1,n) solve(i,a[i]);
	REP(j,1,n) {
		int pos = 1;
		while (!num[j][pos]) ++pos;
		REP(i,pos,M) putchar(num[j][i]+'0');hr;
	}
}

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转载自www.cnblogs.com/uid001/p/10732099.html