题意：给定 n，k，求 while(i <=n) k % i的和。
析：很明显是一个数论题，写几个样例你会发现规律，假设 p = k / i.那么k mod i = k - p*i，如果 k / (i+1) 也是p,那么就能得到 ：
k mod (i+1) = k - p*(i+1) = k mod i - p。所以我们就能得到一个等差数列 k mod (i+1) - k mod i = -p，首项是 p % i。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <ctime>
#include <cstdlib>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){ return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){ return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
LL solve(int a, int d, int n){
return (LL)((LL)n*a - (LL)n*(n-1)/2*d);
}
int main(){
while(scanf("%d %d", &n, &m) == 2){
int i = 1;
LL ans = 0;
while(i <= n){
int a = m % i;
int d = m / i;
int cnt = n - i + 1;
if(d > 0) cnt = Min(cnt, a/d+1);
ans += solve(a, d, cnt);
i += cnt;
}
cout << ans << endl;
}
return 0;
}
题意：给定n, k,求出∑ni=1(k mod i)
