FZU 2104 (13.11.28)
2013-12-03 11:20
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Problem 2104 Floor problem
Accept: 376 Submit: 433
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.
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).
For each test case, print the result of f(n,L)+f(n,L+1)+...+f(n,R) in a single line.
31 2 3100 2 100100 3 100
0382332
“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:
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\n", sum); } return 0; }
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