Educational Codeforces Round 90 (Rated for Div. 2)E. Sum of Digits[数位贡献+思维题+贪心]

题目链接

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题目大意：就是给你$n$和$k$然后再定义一个函数$f(x)是十进制数x各个位数之和$
$叫你求出最小的x使得f(x)+f(x+1)+...+f(x+k)=n|k\in [0,9],n\in [1,150]$

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解题思路：首先我们发现两个规律：
1.就是$x+k$我们最多进位一次
2.你要是想进位的话你要取决俩个东西
1):就是你个位数的大小
2):就是你个位数之后9的个数
3.当上面两个确定之后高位就贪心选
4.还有一个细节就是这$(k+1)$个数的高位应该是一样的
5.那么我们就可以枚举个位数以及个位之后会9的个数

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主函数

 int T;
    read(T);
    while(T --)
    {
        cin >> n >> k;
        ll ans = 1e18;
        _for(i,0,17)//枚举9的个数
         _for(j,0,10)//枚举个位
           ans = min(ans,num(n,k,i,j));
        if(ans == 1e18) ans = -1;
        cout << ans << endl;
    }

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首先我们先讲已经确定的各位数字之和先加上

ll num(int n, int k, int c9, int r) //函数头
{
  int res = 0;
    _for(i,0,k+1) 
    {
        if (i + r < 10) 
        {
            res += i + r + 9 * c9;
        } else {
            res += i + r - 9;
        }
    }

判断一下还剩下数的多少，因为后面的k+1个数的高位都是一样的

int need = n - res;
    if (need % (k + 1) || need < 0)
    {
        return 1e18;
    }

贪心选取每一位

need /= k + 1;
    ll ans = 0;
    if (need < 9)
    {
        ans = need;
    }
    else 
    {
        need -= 8;
        int t = need % 9;
        ans = t;
        for (int i = 1; i <= need / 9; i++) 
        {
            ans = ans * 10 + 9;
        }
        ans = ans * 10 + 8;
    }
   // 这里就是将高位和低位拼接起来
    for (int i = 1; i <= c9; i++) 
    {
        ans = ans * 10 + 9;
    }
    ans = ans * 10 + r;
    return ans;
}

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完整代码

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#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define mid ((l + r) >> 1) 
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define hash Hash
#define next Next
#define count Count
#define pb push_back
#define f first
#define s second
using namespace std;
const int N = 1e6+10, mod = 1e9 + 7;
const double eps = 1e-10;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) 
{
    read(first);
    read(args...);
}
int n, k;
ll num(int n, int k, int c9, int r) 
{
    int res = 0;
    _for(i,0,k+1) 
    {
        if (i + r < 10) 
        {
            res += i + r + 9 * c9;
        } else {
            res += i + r - 9;
        }
    }
    int need = n - res;
    if (need % (k + 1) || need < 0)
    {
        return 1e18;
    }
    need /= k + 1;
    ll ans = 0;
    if (need < 9)
    {
        ans = need;
    }
    else 
    {
        need -= 8;
        int t = need % 9;
        ans = t;
        for (int i = 1; i <= need / 9; i++) 
        {
            ans = ans * 10 + 9;
        }
        ans = ans * 10 + 8;
    }
    for (int i = 1; i <= c9; i++) 
    {
        ans = ans * 10 + 9;
    }
    ans = ans * 10 + r;
    return ans;
}
 
int main()
{
    int T;
    read(T);
    while(T --)
    {
        cin >> n >> k;
        ll ans = 1e18;
        _for(i,0,17)
         _for(j,0,10)
           ans = min(ans,num(n,k,i,j));
        if(ans == 1e18) ans = -1;
        cout << ans << endl;
    }
    return 0;
}