二分查找+大整數加法——Poj 2413 How many Fibs?

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

#define MAXN    500
#define MAXLEN  110
#define LAST MAXLEN-2

char store[MAXN][MAXLEN];   //存儲MAXN個斐波那契數
char *Fibs[MAXN];       //存儲每個斐波那契數的首地址

//大整數相加
char* IntAddition(char *a, char *b, char *sum)
{
    int i, j, k, first;

    //從末位開始，把a與b對應位的和存入sum中，暫不處理進位
    for (i = strlen(a)-1, j = LAST; i >= 0; i--, j--)
    {
        sum[j] = a[i] - '0';
    }
    for (i = strlen(b)-1, k = LAST; i >= 0; i--, k--)
    {
        sum[k] += b[i] - '0';
    }

    //獲取sum中結果的首位位置
    first = j < k ? j : k;

    //處理進位
    for (i = LAST; i >= first; i--)
    {
        sum[i-1] += sum[i] / 10;
        sum[i] = sum[i] % 10 + '0';
    }
    //去除前導'0'
    while (sum[first] == '0' && first < LAST)
    {
        first++;
    }
    //返回sum的首位地址
    return &sum[first];
}

//計算485個斐波那契數
void Fibonacci(void)
{
    memset(store, 0, sizeof(store));
    memset(Fibs, NULL, sizeof(Fibs));

    strcpy(store[1], "1");
    strcpy(store[2], "2");
    Fibs[1] = store[1];
    Fibs[2] = store[2];

    int i;
    for (i = 3; i < 485; i++)
    {
        Fibs[i] = IntAddition(Fibs[i-2], Fibs[i-1], store[i]);
    }
}

int Compare(char *a, char *b)
{
    int lenA = strlen(a);
    int lenB = strlen(b);

    if (lenA == lenB)
    {
        return strcmp(a, b);
    }
    return lenA > lenB ? 1 : -1;
}

int BinarySearch(char *num, bool &flag)
{
    int low = 1;
    int high = 480;

    while (low <= high)
    {
        int mid = (low + high) / 2;

        int res = Compare(num, Fibs[mid]);
        if (res == 0)
        {
            flag = true; return mid;
        }
        else if (res < 0)
        {
            high = mid - 1;
        }
        else
        {
            low = mid + 1;
        }
    }
    return low;
}

int main(void)
{
    Fibonacci();

    char a[MAXLEN], b[MAXLEN];
    while (scanf("%s %s", a, b) != EOF)
    {
        if (strcmp(a, "0") == 0 && strcmp(b, "0") == 0)
        {
            break;
        }

        bool flagLeft = false;
        bool flagRight = false;
        //分別找出a和b在斐波那契數中的位置
        //當查找的數不是斐波那契數時，二分查找返回的位置是第一個比它大的斐波那契數的位置
        int left = BinarySearch(a, flagLeft);
        int right = BinarySearch(b, flagRight);

        //當b也是斐波那契數時，要把兩位置的差值加1
        if (flagRight)
        {
            printf("%d\n", right - left + 1);
        }
        else
        {
            printf("%d\n", right - left);
        }
    }
    return 0;
}