C语言大数据处理

Contents

1. 1.Description
2. 2.Input
3. 3.Output
4. 4.Sample Input
5. 5.Sample Output
6. 6.Source
7. 7.Code
8. 8.note
   1. 8.1.大数处理

Description

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There are many students in PHT School.
 One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. 
 In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM。
 Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7.
  Can you make a program to find the total number of queue with n children?

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)

Output

For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.

Sample Input

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Sample Output

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Source

HDU1297

Code

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//
// Created by Kevin on 2016/10/16.
//
#include<stdio.h>
int main() {
    int n;
    int f[1001][101] = {0};
    f[0][1] = 1;
    f[1][1] = 1;
    f[2][1] = 2;
    f[3][1] = 4;
    for (int i = 4; i < 1001; ++i) {
        for (int j = 1; j < 101; ++j) {
            f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j];    //数组的每一位相加
            f[i][j + 1] += f[i][j] / 10000;    //超过4位的部分保存至数组下一位中
            f[i][j] %= 10000;    //每位数组只保存其中4位
        }
    }
    while (scanf("%d", &n) != EOF) {
        int k = 100;
        while (!f[n][k--]);    //排除前面为空的数组
        printf("%d", f[n][k + 1]);    //输出结果的前四位
        for (; k > 0; --k) {
            printf("%04d", f[n][k]);    //输出其余的所有四位数字，若数字小于四位，则前面用0填充
        }
        printf("\n");
    }
    return 0;
}

note

a.男孩，任何n - 1的合法队列追加1个男孩必然是合法的，情况数为f[n - 1]；

b.女孩，在前n - 1的以女孩为末尾的队列后追加1位女孩也是合法的，我们可以转化为n - 2的队列中追加2位女孩；

一种情况是在n - 2的合法队列中追加2位女孩，情况数为f[n - 2]；

但注意到可能前n - 2位以女孩为末尾的不合法队列（即单纯以1位女孩结尾），也可以追加2位女孩成为合法队列，而这种n - 2不合法队列必然是由n - 4合法队列+1男孩+1女孩的结构，即情况数为f[n - 4]。

大数处理

本题数据量极大，无法用任何数据类型直接存储，于是采用二维数组模拟大数运算，每一位单独相加，每位数组只保存其中4位。