Baskets of Gold Coins

Baskets of Gold Coins

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1862    Accepted Submission(s): 1108

Problem Description

You
are given N baskets of gold coins. The baskets are numbered from 1 to
N. In all except one of the baskets, each gold coin weighs w grams. In
the one exceptional basket, each gold coin weighs w-d grams. A wizard
appears on the scene and takes 1 coin from Basket 1, 2 coins from Basket
2, and so on, up to and including N-1 coins from Basket N-1. He does
not take any coins from Basket N. He weighs the selected coins and
concludes which of the N baskets contains the lighter coins. Your
mission is to emulate the wizard's computation.

Input

The
input file will consist of one or more lines; each line will contain
data for one instance of the problem. More specifically, each line will
contain four positive integers, separated by one blank space. The first
three integers are, respectively, the numbers N, w, and d, as described
above. The fourth 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 be at most 30. The value of d will be less than w.

Output

For
each instance of the problem, your program will produce one line of
output, consisting of one positive integer: the number of the basket
that contains lighter coins than the other baskets.

Sample Input

10 25 8 1109
10 25 8 1045
8000 30 12 959879400

Sample Output

2
10
50

题解：这道题就是说有N个篮子，从1到N-1每个篮子取硬币个数为1.....N-1；

其中有一个篮子的每个硬币价值为w-d，剩余的篮子每个硬币价值为w；

 求这个特殊的篮子取得硬币个数；

思路就是sum是否等于（n-1+1）*(n-1)/2  *w；等于证明这个特殊的篮子没有被取；

代码：

#include<stdio.h>
#include<string.h>
int main(){
    int n,w,d,sum;
    while(~scanf("%d%d%d%d",&n,&w,&d,&sum)){
        int x=w*((n-+)*(n-)/)-sum;
            if(x==)printf("%d\n",n);
            else printf("%d\n",x/d);
    }
    return ;
}