版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/sugarbliss/article/details/89526497
题目链接:https://ac.nowcoder.com/acm/contest/877/K
思路:贪心找到不大于x的数中含有9的数越多越好。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5+7;
ll solve(ll x)
{
ll ans = 0;
while(x)
{
ans += x % 10;
x /= 10;
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
ll x; scanf("%lld",&x);
ll y = 0;
while(y <= x) y = y * 10 + 9;
y /= 10;
printf("%lld\n",solve(y)+solve(x-y));
}
}