codeforces#219_div2_B Making Sequences is Fun
2014-04-06 22:07
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题目地址:cf#219_div2_B
一个很奇怪的数列求和,项数有可能特别多,但是很多相邻的都是一样的,我们一起加上去,这样就不会超时
wa了很多次因为精度,首先取出多少位数就不要用log(k)/log(10); 太容易误差了,用while() 保险。
代码:
一个很奇怪的数列求和,项数有可能特别多,但是很多相邻的都是一样的,我们一起加上去,这样就不会超时
wa了很多次因为精度,首先取出多少位数就不要用log(k)/log(10); 太容易误差了,用while() 保险。
代码:
#include<iostream> #include<cmath> #include<utility> #include<vector> #include<algorithm> using namespace std; typedef pair<int,int> pii; typedef long long inta; inta mypow(inta a,int n) { inta ans=1; for(int i=0;i<n;i++) { ans*=a; } return ans; } int main() { inta w,m,k; cin>>w>>m>>k; inta all=w/k; inta ans=0; // int a=floor(log(m)/log(10)); int a=0; while(mypow(10,a)<=m) { a++; } a--; inta b=mypow(10, a+1);; if(all<=(b-m)*(a+1)) { //cout<<"jingqi"<<endl; ans=all/(a+1); } else { ans=b-m; inta cur=(b-m)*(a+1); int t=0; // cout<<ans<<endl; while(cur<all) { //cout<<cur<<endl; if(all-cur<9LL*b*mypow(10,t)*(a+2+t)) { ans+=(all-cur)/(a+2+t); break; } else { ans+=9LL*b*mypow(10,t); cur+=9LL*b*mypow(10,t)*(a+2+t); //cout<<t<<endl; t++; } } } cout<<ans<<endl; }
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