CodeForces 1060 B Maximum Sum of Digits
Maximum Sum of Digits
You are given a positive integer n.
Let S(x)S(x) be sum of digits in base 10 representation of xx , for example, S(123)=1+2+3=6S(123)=1+2+3=6 , S(0)=0S(0)=0 .
Your task is to find two integers a,ba,b , such that 0≤a,b≤n0≤a,b≤n , a+b=na+b=n and S(a)+S(b)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)(1≤n≤1012) .
Output
Print largest S(a)+S(b)S(a)+S(b) among all pairs of integers a,ba,b , such that 0≤a,b≤n0≤a,b≤n and a+b=na+b=n .
Examples
35
17
10000000000
91
Note
In the first example, you can choose, for example, a=17a=17 and b=18b=18 , so that S(17)+S(18)=1+7+1+8=17S(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=5000000001a=5000000001 and b=4999999999b=4999999999 , with S(5000000001)+S(4999999999)=91S(5000000001)+S(4999999999)=91 . It can be shown that it is impossible to get a larger answer.
解题思路:
给出一个数字n,将他拆分为2个数字,使拆分的两个数字每一位相加的和最大,输出相加后的和。
个人感觉不用管下面提示。我们本着贪心的思想,希望获得尽可能多的9,就是将35,拆分为9与26,将10000000000,拆分为9999999999与1,既将原始数字拆分为比其第一位的全由9组成的数字与另一个补偿数字,补偿数字为原始数字减去拆分的全9数字。之后将拆分的两个数字所有位都相加便可以得到答案。
1 #include<iostream> 2 #include<sstream> 3 #include<string> 4 #include<cstdio> 5 //CodeForces不支持万能头文件bits/stdc++.h 6 using namespace std; 7 typedef long long ll; 8 string n; 9 ll power(int a, int b){ //快速幂 10 ll ans = 1; 11 while(b){ 12 if(b & 1){ 13 ans = ans * a; 14 } 15 a = a * a; 16 b >>= 1; 17 } 18 return ans; 19 } 20 int main() 21 { 22 ll a, b; 23 while(cin >> n) 24 { 25 int suma=0; 26 int sumb=0; 27 a = power(10, n.size() - 1); 28 //若想拆分出小于且9最多的数字,只需要找到n的位数n.size() - 1 29 //之后便可以用10的n.size() - 1次幂找到与n位数相等数字中最小数字 30 //减一便可以得到我们所要拆分的数字 31 istringstream cinn(n); 32 cinn >> b; 33 //先用istringstream读取n中的值输入到整形b中 34 //b - a就是补偿的数字。 35 a--; 36 b = b - a; 37 while(a) //将a的每一位加起来 38 { 39 suma += a % 10; 40 a/=10; 41 } 42 while(b) 43 { 44 sumb += b % 10; //将b的每一位加起来 45 b/=10; 46 } 47 cout << suma + sumb << endl; //所有位数加和 48 } 49 return 0; 50 }