Educational Codeforces Round 5 E. Sum of Remainders

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E. Sum of Remainders

题目链接：点这里

题意：

求n%1+n%2+n%3+...n%m; 答案对1e9+7取模

数据范围 1<n,m<1e13

题解：

我们来看下取模运算是怎么进行的： n%i = n-n/i*i

对原式子化解成 n*m - sum(n/i*i) (1<=i<=m)

然后我们假设n/i~n/j的值是相同的。我们求j的最大值。j=n/(n/j)。

然后就可以把sum(n/i*i)看成若干个等差数列。很容易就能得到答案。

时间复杂度sqrt(n)

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<sstream>#include<algorithm>#include<vector>#include<bitset>#include<set>#include<queue>#include<stack>#include<map>#include<cstdlib>#include<cmath>#define PI 2*asin(1.0)#define LL long long#define pb push_back#define pa pair<int,int>#define clr(a,b) memset(a,b,sizeof(a))#define lson lr<<1,l,mid#define rson lr<<1|1,mid+1,r#define bug(x) printf("%d++++++++++++++++++++%d\n",x,x)#define key_value ch[ch[root][1]][0]const LL  MOD = 1000000000+7;const int N = 5e5+15;const int maxn = 1e6+1000;const int letter = 130;const int INF = 1e17;const double pi=acos(-1.0);const double eps=1e-8;using namespace std;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}LL n,m;LL md(LL x){    return x%MOD;}int main(){    /// n*m-sum(n/i*i)    cin>>n>>m; /// sum = n%1+n%2+n%3...n%m,sum=sum%MOD;    LL ans=0,vs=0;    LL j;    ans=md(md(n)*md(m));    for(LL i=1;i<=min(n,m);i++){        j=n/(n/i);        ///%2        if(j>m) j=m;        LL pc;        if((i+j)&1) pc=md(md((j-i+1)/2)*md(md(n/i)*md(i+j)));        else        pc=md(md(j-i+1)*md(md(n/i)*md((i+j)/2)));        vs=vs+pc;        vs=md(vs);        i=j;    }    ans=md(ans-vs);    if(ans<0) ans+=MOD;    cout << ans << endl;

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