You are given a positive integer
n

Let

S(x) be sum of digits in base 10 representation of
x, for example,
S
(
123

S(123)=1+2+3=6,

S(0)=0.

Your task is to find two integers

a,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
n（1≤n≤1012).

Output
Print largest
S(a)+S(b) among all pairs of integers a,b, such that 0≤a,b≤n and a+b=n.

Examples
inputCopy
35
outputCopy
17
inputCopy
10000000000
outputCopy
91
Note
In the first example, you can choose, for example, 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, witS(5000000001)+S(4999999999)=91. It can be shown that it is impossible to get a larger answer.

要求求一個數(shù)n的兩個可以加和為n的數(shù)的各個位上數(shù)字加和的最大值。。
像這個題，我們要做的就是多出9。。先求不大于n的10的冪次數(shù)，然后再減一。例如：101，不大于它的數(shù)是100，100-1=99。這樣就有了兩個9。思路就是這個。
代碼如下：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std;ll n; ll temp1; ll temp2;int main() {while(scanf("%I64d",&n)!=EOF){int sum1=0;int sum2=0;temp1=1;while(1){temp1*=10;if(temp1*10>=n)break;}temp1--;temp2=n-temp1;while(temp1){sum1+=temp1%10;temp1/=10;}while(temp2){sum2+=temp2%10;temp2/=10;}cout<<sum1+sum2<<endl;}return 0; }

努力加油a啊，(^o)/~

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