10334

其实就是到1000的斐波那契序列，开始开的串的长度200，竟然都能超。。。改成1000就对了。

稍微的难点就是用到大数加法。

//============================================================================
// Name        : 10334.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;


int n, t;
long long a, b, c;
char sa[1000], sb[1000], sc[1000];

//void add(char *a, char *b)
//{
//    int lena = strlen(a);
//    int lenb = strlen(b);
//    int len = lena > lenb ? lena : lenb;
//    len++;
//    memset(a+lena, '0', len-lena);
//    memset(b+lenb, '0', len-lenb);
//    int c, t;
//    c = 0;
//    for(int i = 0;i < len;i++)
//    {
//        t = a[i] - '0' + b[i] - '0' + c;
//        a[i] = t%10+'0';
//        c = t/10;
//    }
//    if(a[len-1] == '0')
//    {
//        a[len-1] = 0;
//    }
//    strcpy(sc, a);
//}

void add(char *a, char *b){
	int len1, len2, sans[1000];
	char s1[1000], s2[1000], st[1000];
	len1 = strlen(a);
	len2 = strlen(b);
	if(len1 > len2){
		strcpy(st, a);
		strcpy(a, b);
		strcpy(b, st);
	}
	len1 = strlen(a);
	len2 = strlen(b);
	for(int i = 0;i < len1;i++){
		s1[i] = a[len1-1-i];
	}
	s1[len1] = '\0';
	for(int i = 0;i < len2;i++){
		s2[i] = b[len2-1-i];
	}
	s2[len2] = '\0';
	c = 0;
	for(int i = 0;i < len2;i++){
		if(i < len1){
			sans[i] = (s1[i]-'0'+s2[i]-'0'+ c)%10;
			c = (s1[i]-'0'+s2[i]-'0'+ c)/10;
		}
		else{
			sans[i] = (s2[i]-'0'+c)%10;
			c = (s2[i]-'0'+c)/10;
		}
	}
	if(c != 0){
		len2 += 1;
		sans[len2-1] = c;
	}
	for(int i = 0;i < len2;i++){
		sc[i] = sans[len2-1-i]+'0';
	}
	sc[len2] = '\0';
}


int main() {
	freopen("a.txt", "r", stdin);
	while(scanf("%d", &n)!=EOF){
		if(n == 0){
			printf("1\n");
			continue;
		}
		if(n == 1){
			printf("2\n");
			continue;
		}
		strcpy(sa, "1");
		strcpy(sb, "2");
		for(int i = 2;i <= n;i++){
			add(sa, sb);
			strcpy(sa, sb);
			strcpy(sb, sc);
		}
		printf("%s\n", sc);
	}
	return 0;
}