CodeForces 468 C.Hack it!(数学)
2017-07-20 09:17
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Description
定义f(n)为n的十进制表示各位数之和,求一组满足
的l和r
Input
一个正整数a(1<=a<=1e18)
Output
输出一组满足条件的l和r(1<=l<=r<=1e200)
Sample Input
46
Sample Output
1 10
Solution
令sum(l,r)=f(l)+f(l+1)+…+f(r)
注意到f(x+1e18)=f(x)+1,故sum(x+1,x+1e18)=sum(x,x+1e18-1)+1
如果sum(0,1e18-1)%a=k,则把该区间右移a-k即得到一组解
sum(0,9)=45
sum(0,99)=45*10+450=900
sum(0,999)=900*10+4500=13500
……
sum(0,10^i-1)=sum(0,10^(i-1)-1)*10+45*10^(i-1)
……
sum(0,1e18-1)=81*1e18
Code
定义f(n)为n的十进制表示各位数之和,求一组满足
的l和r
Input
一个正整数a(1<=a<=1e18)
Output
输出一组满足条件的l和r(1<=l<=r<=1e200)
Sample Input
46
Sample Output
1 10
Solution
令sum(l,r)=f(l)+f(l+1)+…+f(r)
注意到f(x+1e18)=f(x)+1,故sum(x+1,x+1e18)=sum(x,x+1e18-1)+1
如果sum(0,1e18-1)%a=k,则把该区间右移a-k即得到一组解
sum(0,9)=45
sum(0,99)=45*10+450=900
sum(0,999)=900*10+4500=13500
……
sum(0,10^i-1)=sum(0,10^(i-1)-1)*10+45*10^(i-1)
……
sum(0,1e18-1)=81*1e18
Code
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue> #include<map> #include<set> #include<ctime> using namespace std; typedef long long ll; #define INF 0x3f3f3f3f #define maxn 1111 ll a,m=1e18; int main() { while(~scanf("%I64d",&a)) { ll ans=a-m%a*9%a*9%a; printf("%I64d %I64d\n",ans,ans+m-1); } return 0; }
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