Description
You are given a positive integer nn.
Let S(x) be sum of digits in base 10 representation of xx, for example, S(123)=1+2+3=6, S(0)=0.
Your task is to find two integers a,ba,b, such that 0≤a,b≤n, a+b=n and S(a)+S(b) is the largest possible among all such pairs.
Input
The only line of input contains an integer nn (1≤n≤1012).
Output
Print largest S(a)+S(b) among all pairs of integers a,ba,b, such that 0≤a,b≤n and a+b=n.
Sample Input
35
17
10000000000
91
Hint
In the first example, you can choose, for example, a=17 and b=18, so that S(17)+S(18)=1+7+1+8=17. It can be shown that it is impossible to get a larger answer.
In the second test example, you can choose, for example, a=5000000001 and b=4999999999, with S(5000000001)+S(4999999999)=91. It can be shown that it is impossible to get a larger answer.
相信不少人wa是这个代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e 2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std; int main()
{
ll n,a,b;
cin>>n;
ll m=n,k=;
while(m>=k)
{
k=k*+;
}
k/=;
//cout<<k<<endl;
a=k;
b=n-k;
ll sum=;
while(a)
{
sum+=a%;
a/=;
}
while(b)
{
sum+=b%;
b/=;
}
cout<<sum<<endl;
return ;
}
你们可以尝试一下123这个样例，输出为15,但正确答案为24（99+24）。
AC代码如下：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e 2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std; int main()
{
ll n,a,b;
cin>>n;
ll m=n,k=;
while(m>=k)
{
k=k*+;
}
k/=;
//cout<<k<<endl;
a=k;
b=n-k;
ll sum=;
while(a)
{
sum+=a%;
a/=;
}
while(b)
{
sum+=b%;
b/=;
}
cout<<sum<<endl;
return ;
}