UVA10407 Simple division【同余+一阶差分】

Integer division between a dividend n and a divisor d yields a quotient q and a remainder r. q is the integer which maximizes q ∗ d such that q ∗ d ≤ n and r = n − q ∗ d.
For any set of integers there is an integer d such that each of the given integers when divided by d leaves the same remainder.

Input
Each line of input contains a sequence of nonzero integer numbers separated by a space. The last number on each line is 0 and this number does not belong to the sequence. There will be at least 2 and no more than 1000 numbers in a sequence; not all numbers occuring in a sequence are equal. The last line of input contains a single 0 and this line should not be processed.
Output
For each line of input, output the largest integer which when divided into each of the input integers leaves the same remainder.
Sample Input
701 1059 1417 2312 0
14 23 17 32 122 0
14 -22 17 -31 -124 0
0
Sample Output
179
3
3

问题链接：UVA10407 Simple division
问题简述：给定若干个整数（这些数不完全相同），求使这些数同余的最大除数。
问题分析：（略）
先求差值，然后再做GCD运算。
给出2个解法程序，不用数组存储的解法要好一些。
程序说明：（略）
参考链接：（略）
题记：不使用数组存储才是真功夫。

AC的C++语言程序如下：

/* UVA10407 Simple division */#include <bits/stdc++.h>using namespace std;int main()
{int a1, a;while(~scanf("%d", &a1) && a1) {int g = 0;while(scanf("%d", &a) == 1 && a) {int d = a - a1;if(d) {if(g) g = __gcd(g, d); else g = d;}a1 = a;}printf("%d\n", abs(g));}return 0;
}

AC的C++语言程序如下：

/* UVA10407 Simple division */#include <bits/stdc++.h>using namespace std;const int N = 1000;
int d[N];int main()
{int a1, a, cnt, k;while(~scanf("%d", &a1) && a1) {for(cnt = 0; scanf("%d", &a) == 1 && a; cnt++)d[cnt] = a - a1, a1 = a;for(k = 0; d[k] == 0; k++);int g = d[k++];for(; k < cnt; k++)if(d[k]) g = __gcd(g, d[k]);printf("%d\n", abs(g));}return 0;
}

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UVA10407 Simple division【同余+一阶差分】

UVA10407 Simple division【同余+一阶差分】相关推荐

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