Harbour.Space Scholarship Contest 2021-2022 (open for everyone, rated, Div. 1 + Div. 2)
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Harbour.Space Scholarship Contest 2021-2022 (open for everyone, rated, Div. 1 + Div. 2)
题号 | 题目 | 知识点 |
---|---|---|
A | Digits Sum | 签到 |
B | Reverse String | |
C | Penalty | |
D | Backspace | |
E | Permutation Shift | |
F | Pairwise Modulo | |
G | Common Divisor Graph | |
H | XOR and Distance | |
I | Stairs |
cf1553 A. Digits Sum
题意:
定义S(x):为x的各位之和
问1<=x<=n,存在多少S(x+1)<S(x)?
题解:
不难看出满足这个情况的x末尾肯定是9,而且仔细观察输出,貌似答案就是把最后一位抹去,但也不完全是,因为如果最后一位是9,它本身也是,所以答案是(x+1)/10
代码:
#include<bits/stdc++.h>
#define debug(a,b) printf("%s = %d\\n",a,b);
typedef long long ll;
using namespace std;
//qdu打铁匠
inline int read(){
int s=0,w=1;
char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();//s=(s<<3)+(s<<1)+(ch^48);
return s*w;
}
int main()
{
int t;
cin>>t;
while(t--){
int n;
cin>>n;
n++;
cout<<n/10<<endl;
}
}
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