HDU 2401 Baskets of Gold Coins

http://acm.hdu.edu.cn/showproblem.php?pid=2401

 

Baskets of Gold Coins

Time Limit: 1000/1000 MS(Java/Others)    MemoryLimit: 32768/32768 K (Java/Others)
Total Submission(s):700    AcceptedSubmission(s): 352

Problem Description

You are given N baskets of gold coins.The baskets are numbered from 1 to N. In all except one of thebaskets, each gold coin weighs w grams. In the one exceptionalbasket, each gold coin weighs w-d grams. A wizard appears on thescene and takes 1 coin from Basket 1, 2 coins from Basket 2, and soon, up to and including N-1 coins from Basket N-1. He does not takeany coins from Basket N. He weighs the selected coins and concludeswhich of the N baskets contains the lighter coins. Your mission isto emulate the wizard's computation.

 

 

Input

The input file will consist of one ormore lines; each line will contain data for one instance of theproblem. More specifically, each line will contain four positiveintegers, separated by one blank space. The first three integersare, respectively, the numbers N, w, and d, as described above. Thefourth integer is the result of weighing the selected coins.

N will be at least 2 and not more than 8000. The value of w will beat most 30. The value of d will be less than w.

 

 

Output

For each instance of the problem, yourprogram will produce one line of output, consisting of one positiveinteger: the number of the basket that contains lighter coins thanthe other baskets.

 

 

Sample Input

10 25 81109 10 25 8 1045 8000 30 12 959879400

 

 

Sample Output

2 1050

 

 

Source

ACM/ICPC 2008 Warmup（2）——测试帐号（杭州）

 

 

Recommend

lcy

 

题意：有N个篮子，编号1—N，篮子中有很多金币，每个重w.但是有一个编号的篮子中，每个金币重d.现从第一个篮子中拿1个金币，第二个篮子中拿2个……第N-1中拿N-1个，第N中不拿，给出这些金币的总重量wei，问：是第几个篮子中的金币重量较轻？

分析：一道数学题，先求1—N篮子金币应有的总重量w*（1+n-1)(n-1)/2-wei等差数列求和，再乘每个金币应有的重量，所求和减去wei得到金币重量的差值。若为0，则必为编号N；若不为0，除以d，得到较轻金币的个数，即为所求编号。

代码：

#include<stdio.h>
int main()
{
 int n,w,d,wei,ans;
 while(scanf("%d%d%d%d",&n,&w,&d,&wei)==4)
 {
   ans=w*n*(n-1)/2-wei;
   ans=ans/d;
   if(ans==0)printf("%d\n",n);
   elseprintf("%d\n",ans); 
 }
 return 0;
}