【链接】

【题意】

让你求出所有x的和

其中

(x div b)是(x mod b)的倍数

且x mod b不等于0

且(x div b)除(x mod b)的值(假设为k),k∈[1..a]

【题解】

枚举k从1~a

这样k就固定了

n = x / b;

m = x % b;

n = mk ····①

x = bn + m

x = (bk+1)m

而m∈[1..b-1]

所以求和一下就是∑x = (bk+1)( b*(b-1)/2)

这里我们m属于1..b-1,显然由①式可知,k不同的时候,m从1到b-1变化的话,

得到的n也是不同的.(因此每一组(n,m)都不会相同)

所以如果我们枚举不同的k值,然后m从1到b-1变化不会得到重复的x值的,因为每一个x值

都唯一标识了一组(n,m),而(n,m)不会重复

枚举k加起来就好啦

【代码】

import java.io.*;import java.util.*;public class Main {            static InputReader in;    static PrintWriter out;            public static void main(String[] args) throws IOException{        //InputStream ins = new FileInputStream("E:\\rush.txt");        InputStream ins = System.in;        in = new InputReader(ins);        out = new PrintWriter(System.out);        //code start from here        new Task().solve(in, out);        out.close();    }        static int N = 50000;    static class Task{        /*         * n = x / b;         * m = x % b;         * n = m*k         * x = b*n + m         * x = (b*k+1)*m         * m 1..b-1         * ∑x = (b*k+1)*( b*(b-1)/2)         * k 1..a         */        long a,b;        long MOD = (int)(1e9+7),ans = 0;                public void solve(InputReader in,PrintWriter out) {            a = in.nextInt();b = in.nextInt();                        for (long k = 1;k <= a;k++) {                long temp = (b*k+1)%MOD;                long temp1 = b*(b-1)/2%MOD;                temp = temp*temp1%MOD;                ans = (ans+temp)%MOD;            }            out.println(ans);        }    }        static class InputReader{        public BufferedReader br;        public StringTokenizer tokenizer;                public InputReader(InputStream ins) {            br = new BufferedReader(new InputStreamReader(ins));            tokenizer = null;        }                public String next(){            while (tokenizer==null || !tokenizer.hasMoreTokens()) {                try {                tokenizer = new StringTokenizer(br.readLine());                }catch(IOException e) {                    throw new RuntimeException(e);                }            }            return tokenizer.nextToken();        }                public int nextInt() {            return Integer.parseInt(next());        }    }}