Sum of Remainders(数学题)
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F - Sum of Remainders
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
Calculate the value of the sum: n mod 1 + n mod 2 + n mod 3 + ... + n mod m. As the result can be very large, you should print the value modulo 109 + 7 (the remainder when divided by 109 + 7).
The modulo operator a mod b stands for the remainder after dividing a by b. For example 10 mod 3 = 1.
Input
The only line contains two integers n, m (1 ≤ n, m ≤ 1013) — the parameters of the sum.
Output
Print integer s — the value of the required sum modulo 109 + 7.
Sample Input
3 4
4
4 4
1
1 1
0
//给你两个数n,m,问你n % 1 + n % 2 + … + n% m为几
这个题目的思路是,和为 n * m - sum ( [ n / i ] * i ) ,[ ] 是向下取整,i 从(1--- m)
具体是:
前面的 n*m 肯定就这样了
主要是后面的 :将 [ n / i ] 相同的做一个区间,用求和公式去节省时间
i 从 1 --- sqrt (n) ;
l = n / ( i + 1 ) + 1 , r = n / i ; // 这就是一个个的区间
比如 n = 20 , m = 20
i=1 --> l=11, r=20 n / (11---20) 都是等于 1
i=2 --> l=7, r=10 n / (7---10) 都等于2
i=3 --> l=r=6
i=4 --> l=r=5
注意一些特殊情况,看注释
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 using namespace std; 5 6 #define ll __int64 7 const ll MOD=1e9+7; 8 9 int main() 10 { 11 ll n,m; 12 scanf("%I64d%I64d",&n,&m); 13 ll ans=(n%MOD)*(m%MOD)%MOD; 14 ll temp=0,las=m+1;//记录哪些数没被减 15 m=min(n,m);//n 余大于 n 的都等于 n 16 ll nn=(ll)sqrt(n*1.0); 17 for (ll i=1;i<=nn;i++) 18 { 19 ll l = n/(i+1)+1; 20 ll r = n/i; 21 22 r=min(r,m);//可能 r 比 m 大 23 if (l>r) continue; 24 25 las=min(las,l); 26 27 ll s1=l+r , s2 =(r-l+1);//这里高斯求和有个坑,要先判断哪个数可以除2,再乘 28 if (s1%2==0) s1/=2; //直接用公式也不对,会超出ll限制 29 else s2/=2; 30 s1%=MOD;s2%=MOD; 31 s1=(s1*s2)%MOD; 32 s1=s1*i%MOD; 33 temp=(temp+s1)%MOD; 34 } 35 ans=(ans+MOD-temp)%MOD; 36 for (ll i=1;i<las;i++) //剩下的没被减得数,将n余之为0的最大整数减去 37 { 38 temp=n/i%MOD*i%MOD; 39 ans=(ans+MOD-temp)%MOD; 40 } 41 printf("%I64d\\n",ans); 42 43 return 0; 44 }
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