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  • FZU 2104 (13.11.28)

    Problem 2104 Floor problem

    Accept: 376    Submit: 433
    Time Limit: 1000 mSec    Memory Limit : 32768 KB

    Problem Description

    In this problem, we have f(n,x)=Floor[n/x]. Here Floor[x] is the biggest integer such that no larger than x. For example, Floor[1.1]=Floor[1.9]=1, Floor[2.0]=2.

    You are given 3 positive integers n, L and R. Print the result of f(n,L)+f(n,L+1)+...+f(n,R), please.

    Input

    The first line of the input contains an integer T (T≤100), indicating the number of test cases.

    Then T cases, for any case, only 3 integers n, L and R (1≤n, L, R≤10,000, L≤R).

    Output

    For each test case, print the result of f(n,L)+f(n,L+1)+...+f(n,R) in a single line.

    Sample Input

    31 2 3100 2 100100 3 100

    Sample Output

    0382332

    Source

    “高教社杯”第三届福建省大学生程序设计竞赛


    为高教杯复习而做,水题

    直接贴AC代码:
    #include<stdio.h>
    
    int f(double x) {
        int i;
        for(i = 0; i <= 10000; i++)
            if(x >= i && x < i+1)
                break;
        return i;
    }
    
    int main() {
        int T;
        scanf("%d", &T);
        while(T--) {
            int n, l, r;
            int sum = 0;
            scanf("%d %d %d", &n, &l, &r);
            for(int i = l; i <= r; i++) {
                double num = n / i;
                sum += f(num);
            }
            printf("%d
    ", sum);
        }
        return 0;
    }

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  • 原文地址:https://www.cnblogs.com/fuhaots2009/p/3455397.html
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