HDU 1316 How Many Fibs?

 

http://acm.hdu.edu.cn/showproblem.php?pid=1316

题目大意：给你一个区间，[a,b] ， a <= b <= 10^100，叫你统计在这个区间内有多少个的斐波那契数。

解题思路：注意到a <= b <= 10^100 ，也就是说我们只要打表打到某个斐波那契数的长度为 100 位的表就好，之后给定任何的区间都可以通过搜索表来统计。通过打表，我们可以知道，我们要求的最大的斐波那契数为第 518 个数，所以数组就只要开 520 就好了，剩下的就是大数问题了，大家自己可以解决的啦。

 

#include <stdio.h> #include <string.h> #define size 519 #define LL __int64 const unsigned LL WIDTHMAX=1000000000; const unsigned int WIDTH = 9; struct BigInt { LL val[15]; int len; BigInt(){len=0;} }fib[size]; /*字符串转换为大数*/ BigInt StrToBigInt(char *s) { BigInt a; a.len = 0; int i = strlen(s); int j; __int64 sum = 0; while ( i>=WIDTH) { for (j=i-WIDTH; j<i; j++) sum = sum * 10 + (s[j] - '0'); a.val[a.len++] = sum; sum = 0; i -= WIDTH; } if (i > 0) { for (int j=0; j<i; j++) sum = sum * 10 + (s[j] - '0'); a.val[a.len++] = sum; } return a; } /*大数相加*/ BigInt AddBigInt(const BigInt & a, const BigInt & b) { //逆序计算a+b，则从低位开始计算 BigInt c; LL up = 0; LL i = 0; c.len = 0; while (i<a.len&&i<b.len) { c.val[c.len++] = (a.val[i] + b.val[i] + up) % WIDTHMAX; up = (a.val[i] + b.val[i] + up) / WIDTHMAX; i++; } while (i<a.len) { c.val[c.len++] = (a.val[i] + up) % WIDTHMAX; up = (a.val[i] + up) / WIDTHMAX; i++; } while (i<b.len) { c.val[c.len++] = (b.val[i] + up) % WIDTHMAX; up = (b.val[i] + up) / WIDTHMAX; i++; } if (up>0) c.val[c.len++] = up; return c; } /*比较大小*/ int ComPareBigInt(const BigInt & a, const BigInt & b) { int i; if (a.len > b.len) return 1; if (a.len < b.len) return -1; for (i=a.len-1; i>=0; i--) { if (a.val[i] > b.val[i]) return 1; if (a.val[i] < b.val[i]) return -1; } return 0; /*相等*/ } int main() { int i,count; char c[101],d[101]; BigInt a,b; fib[1].val[0]=fib[1].len=fib[2].len=1; fib[2].val[0]=2; /*打表*/ for (i=3;;i++) { fib[i] = AddBigInt(fib[i-1],fib[i-2]); if(fib[i].len>=13)/*此时i = 518*/ break; } while (scanf("%s%s",c,d)!=EOF) { if(strlen(c)==1&&strlen(d)==1&&c[0]=='0'&&d[0]=='0') break; count = 0; a = StrToBigInt(c); b = StrToBigInt(d); for (i=1;;i++) { if(ComPareBigInt(fib[i],a)<0)/*小于左边界*/ continue; if (ComPareBigInt(a,fib[i])<=0&&ComPareBigInt(fib[i],b)<=0)/*夹在两个数之间则统计*/ count++; else break; } printf("%d/n",count); } return 0; }