BZOJ 1257: [CQOI2007]余数之和sum
1257: [CQOI2007]余数之和sum
Time Limit: 5 Sec Memory Limit: 162 MBSubmit: 3769 Solved: 1734
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Description
给出正整数n和k,计算j(n, k)=k mod 1 + k mod 2 + k mod 3 + … + k mod n的值,其中k mod i表示k除以i的余数。例如j(5, 3)=3 mod 1 + 3 mod 2 + 3 mod 3 + 3 mod 4 + 3 mod 5=0+1+0+3+3=7
Input
输入仅一行,包含两个整数n, k。
Output
输出仅一行,即j(n, k)。
Sample Input
5 3
Sample Output
7
HINT
50%的数据满足:1<=n, k<=1000 100%的数据满足:1<=n ,k<=10^9
Source
分析:
最近学习数学...先写道水题压压惊TAT...
Σ(1<=i<=n) k%i
=Σ(1<=i<=n) k-(k/i)*i
=n*k-Σ(1<=i<=n) (k/i)*i
发现k/i的值不超过sqrt(n)种,可以分段计算,而且貌似n>k的时候答案是固定的...
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 #include<cmath> 6 //by NeighThorn 7 using namespace std; 8 //大鹏一日同风起,扶摇直上九万里 9 10 int n,k,m; 11 12 long long ans; 13 14 signed main(void){ 15 scanf("%d%d",&n,&k); 16 ans=(long long)n*(long long)k; 17 if(n>k)n=k; 18 for(int i=1,l,r,x;i<=n;i=r+1){ 19 x=k/i;r=k/x,l=k/(x+1)+1; 20 if(r>n) 21 r=n; 22 ans-=(long long)(r+l)*(long long)(r-l+1)*x/2; 23 } 24 printf("%lld\n",ans); 25 return 0; 26 }
by NeighThorn