本文主要是介绍CF Infinite Sequence，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Infinite Sequence
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
描述
Vasyalikes everything infinite. Now he isstudying the properties of a sequence s, such that its first element is equalto a (s1 = a), and the difference between any two neighbouring elements isequal to c (si - si - 1 = c). Inparticular, Vasya wonders if his favourite integer b appears in this sequence,that is, there exists a positive integer i, such that si = b. Of course, you are the person he asks for ahelp.
Vasya喜欢一切无限。现在,他正在研究一个序列s的属性,这样它的第一个元素等于a ,并且和任何两个相邻元素之间的区别=c(si - si - 1 = c)。特别是Vasya感兴趣他最喜欢整数b出现在这个序列,也就是说,存在一个正整数,这样如果=b。当然,你是他要求帮助的人。
Input
输入
Thefirst line of the input contain three integers a, b and c ( - 109 ≤ a, b, c ≤ 109)—the first element of the sequence, Vasya's favorite number and the differencebetween any two neighbouring elements of the sequence, respectively.
输入的第一行包含三个整数a,b和c(-109≤a,b,c≤109)-第一个元素的序列,Vasya最喜欢的数量和序列的任何两个相邻元素之间的区别,分别。
Output
输出
Ifb appears in the sequence s print "YES" (without quotes), otherwiseprint "NO" (without quotes).
如果b出现在序列s打印“YES”(没有引号),否则打印“不”(没有引号)。
Sample Input
Input
17 3
Output
YES
Input
1010 0
Output
YES
Input
1-4 5
Output
NO
Input
060 50
Output
NO
题意：就是给定三个数a,b,c。判断能不能满足a+n*c==b
思路：1，考虑特殊情况
2，利用（b-a）/a的整型与浮点型答案比较，相等就是YES。另外 n >=0
/*=============================AC情况===============================*/
/*题目网址: */
/*时间： */
/*心得： */#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define G 100int main() {int sign1;double sign2,a,b,c;while(scanf("%lf%lf%lf",&a,&b,&c)!=EOF) {if(c==0) {if(a==b)printf("YES\n");elseprintf("NO\n");} else {sign1=(int)((b-a)/c);sign2=((b-a)/c);//printf("%d %lf",sign1,sign2);if(sign1>=0&&sign1==sign2)printf("YES\n");elseprintf("NO\n");}}return 0;
}/*********************************测试数据***********************************************************************************************************/
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