Maximum Sum of Digits（CodeForces 1060B）
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
- 1 #include<iostream>
- 2 #include<cstdio>
- 3 #include<cstdlib>
- 4 #include<cstring>
- 5 #include<string>
- 6 #include<cmath>
- 7 #include<map>
- 8 #include<stack>
- 9 #include<vector>
- 10 #include<queue>
- 11 #include<set>
- 12 #include<algorithm>
- 13 #define max(a,b) (a>b?a:b)
- 14 #define min(a,b) (a<b?a:b)
- 15 #define swap(a,b) (a=a+b,b=a-b,a=a-b)
- 16 #define maxn 320007
- 17 #define N 100000000
- 18 #define INF 0x3f3f3f3f
- 19 #define mod 1000000009
- 20 #define e 2.718281828459045
- 21 #define eps 1.0e18
- 22 #define PI acos(-1)
- 23 #define lowbit(x) (x&(-x))
- 24 #define read(x) scanf("%d",&x)
- 25 #define put(x) printf("%d\n",x)
- 26 #define memset(x,y) memset(x,y,sizeof(x))
- 27 #define Debug(x) cout<<x<<" "<<endl
- 28 #define lson i << 1,l,m
- 29 #define rson i << 1 | 1,m + 1,r
- 30 #define ll long long
- 31 //std::ios::sync_with_stdio(false);
- 32 //cin.tie(NULL);
- 33 using namespace std;
- 34
- 35
- 36 int main()
- 37 {
- 38 ll n,a,b;
- 39 cin>>n;
- 40 ll m=n,k=9;
- 41 while(m>=k)
- 42 {
- 43 k=k*10+9;
- 44 }
- 45 k/=10;
- 46 //cout<<k<<endl;
- 47 a=k;
- 48 b=n-k;
- 49 ll sum=0;
- 50 while(a)
- 51 {
- 52 sum+=a%10;
- 53 a/=10;
- 54 }
- 55 while(b)
- 56 {
- 57 sum+=b%10;
- 58 b/=10;
- 59 }
- 60 cout<<sum<<endl;
- 61 return 0;
- 62 }
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：
- 1 #include<iostream>
- 2 #include<cstdio>
- 3 #include<cstdlib>
- 4 #include<cstring>
- 5 #include<string>
- 6 #include<cmath>
- 7 #include<map>
- 8 #include<stack>
- 9 #include<vector>
- 10 #include<queue>
- 11 #include<set>
- 12 #include<algorithm>
- 13 #define max(a,b) (a>b?a:b)
- 14 #define min(a,b) (a<b?a:b)
- 15 #define swap(a,b) (a=a+b,b=a-b,a=a-b)
- 16 #define maxn 320007
- 17 #define N 100000000
- 18 #define INF 0x3f3f3f3f
- 19 #define mod 1000000009
- 20 #define e 2.718281828459045
- 21 #define eps 1.0e18
- 22 #define PI acos(-1)
- 23 #define lowbit(x) (x&(-x))
- 24 #define read(x) scanf("%d",&x)
- 25 #define put(x) printf("%d\n",x)
- 26 #define memset(x,y) memset(x,y,sizeof(x))
- 27 #define Debug(x) cout<<x<<" "<<endl
- 28 #define lson i << 1,l,m
- 29 #define rson i << 1 | 1,m + 1,r
- 30 #define ll long long
- 31 //std::ios::sync_with_stdio(false);
- 32 //cin.tie(NULL);
- 33 using namespace std;
- 34
- 35
- 36 int main()
- 37 {
- 38 ll n,a,b;
- 39 cin>>n;
- 40 ll m=n,k=9;
- 41 while(m>=k)
- 42 {
- 43 k=k*10+9;
- 44 }
- 45 k/=10;
- 46 //cout<<k<<endl;
- 47 a=k;
- 48 b=n-k;
- 49 ll sum=0;
- 50 while(a)
- 51 {
- 52 sum+=a%10;
- 53 a/=10;
- 54 }
- 55 while(b)
- 56 {
- 57 sum+=b%10;
- 58 b/=10;
- 59 }
- 60 cout<<sum<<endl;
- 61 return 0;
- 62 }
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