二分查找+大整数加法——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];   //存储MAXN个斐波那契数
char *Fibs[MAXN];       //存储每个斐波那契数的首地址

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

//从末位开始，把a与b对应位的和存入sum中，暂不处理进位
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';
}

//获取sum中结果的首位位置
first = j < k ? j : k;

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

//计算485个斐波那契数
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;
//分别找出a和b在斐波那契数中的位置
//当查找的数不是斐波那契数时，二分查找返回的位置是第一个比它大的斐波那契数的位置
int left = BinarySearch(a, flagLeft);
int right = BinarySearch(b, flagRight);

//当b也是斐波那契数时，要把两位置的差值加1
if (flagRight)
{
printf("%d\n", right - left + 1);
}
else
{
printf("%d\n", right - left);
}
}
return 0;
}