zoukankan      html  css  js  c++  java
  • 九度OJ 1442 A sequence of numbers

    题目地址:http://ac.jobdu.com/problem.php?pid=1442

    题目描述:

    Xinlv wrote some sequences on the paper a long time ago, they might be arithmetic or geometric sequences. The numbers are not very clear now, and only the first three numbers of each sequence are recognizable. Xinlv wants to know some numbers in these sequences, and he needs your help.

    输入:

    The first line contains an integer N, indicting that there are N sequences. Each of the following N lines contain four integers. The first three indicating the first three numbers of the sequence, and the last one is K, indicating that we want to know the K-th numbers of the sequence.

    You can assume 0 < K <= 10^9, and the other three numbers are in the range [0, 2^63). All the numbers of the sequences are integers. And the sequences are non-decreasing.

    输出:

    Output one line for each test case, that is, the K-th number module (%) 200907.

    样例输入:
    2
    1 2 3 5
    1 2 4 5
    样例输出:
    5
    16
    #include <stdio.h>
     
    #define M 200907
     
    long long fun1(long long data[], int k){
        long long p = data[1] - data[0];
        return ((data[0] % M) + (((k-1) % M) * (p % M)) % M) % M;
    }
     
    long long fun2(long long data[], int k){
        long long p = data[1] / data[0];
        long long ans = data[0];
        --k;
        while (k != 0){
            if (k % 2 == 1){
                ans = (ans * p) % M;
            }
            k /= 2;
            p = (p * p) % M;
        }
        return ans;
    }
     
    long long KthNumber(long long data[], int k){
        int flag;
     
        if ((data[1] - data[0]) == (data[2] - data[1]))
            flag = 0;
        else
            flag = 1;
        if (flag == 0){
            return fun1(data, k);
        }
        else{
            return fun2(data, k);
        }
    }
     
    int main(void){
        int n;
        long long data[3];
        int k;
        int i;
     
        while (scanf ("%d", &n) != EOF){
            while (n-- != 0){
                for (i=0; i<3; ++i){
                    scanf ("%lld", &data[i]);
                }
                scanf ("%d", &k);
                printf ("%d
    ", KthNumber (data, k));
            }
        }
     
        return 0;
    }


  • 相关阅读:
    分布式日志收集系统: Facebook Scribe之日志收集方案
    20111030 19:37 杨辉三角形 (java)
    pku acm 1833 排列
    俞敏洪郑大演讲经典语句
    自己在inode客户端的大量问题(不断更新中)(20120223 21:24 )
    智力测验:硬币问题
    windows up可以更新但是无法上网的一天挣扎
    hdu1754 I Hate It
    acm算法资源网站
    pku3041 Asteroids
  • 原文地址:https://www.cnblogs.com/liushaobo/p/4373784.html
Copyright © 2011-2022 走看看