Maximum Sum of Digits

最新推荐文章于 2021-03-24 19:02:11 发布

newhonor

最新推荐文章于 2021-03-24 19:02:11 发布

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CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/weixin_39132605/article/details/89523789

本文探讨了在给定整数n的情况下，寻找两个整数a和b，使得a+b=n且a和b的数字和最大的问题。通过预处理9的幂次序列，算法能够快速找到接近n的最大数字，然后计算其数字和。

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https://ac.nowcoder.com/acm/contest/877/K

题意：n；a+b=n，找a，b使得他们的数字和最大；

思路：处理除9,99,999,9999....减去最大的算就行，一开始处理9,19,199...脑抽:(

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdio>
#include<map>
#include<stack>
#include<string>
#include<bits/stdc++.h>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=1e6+9;
const int mod=1e9+7;

inline ll read()
{
    ll f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9')
    {
        if(ss=='-')f=-1;ss=getchar();
    }
    while(ss>='0'&&ss<='9')
    {
        x=x*10+ss-'0';ss=getchar();
    }    return f*x;
}

ll num[100];
int main()
{
    //FAST_IO;
    //freopen("input.txt","r",stdin);

    int cnt=0;
    num[cnt++]=9;
    ll now=9;
    for(int i=1;i<14;i++)
    {
        num[cnt++]=now*10+9;
        now=now*10+9;
    }
    //for(int i=0;i<cnt;i++) cout<<num[i]<<" ";
    int t;
    cin>>t;
    while(t--)
    {
        ll n;
        ll ans=0;
        cin>>n;
        int pos=lower_bound(num,num+cnt,n)-num;
        if(num[pos]>n)pos--;
        ll k=n-num[pos];
       // cout<<num[pos]<<" "<<k<<endl;
        while(k)
        {
            ans=ans+k%10;
            k/=10;
        }
        ll tmp=num[pos];
        while(tmp)
        {
            ans+=tmp%10;
            tmp/=10;
        }
        cout<<ans<<endl;
    }
    return 0;
}