hdu4099 Revenge of Fibonacci

Posted 2020-08-10 曹孟德

   题意：给定fibonacci数列，输入前缀，求出下标。题目中fibonacci数量达到100000，而题目输入的前缀顶多为40位数字，这说明我们只需要精确计算fibinacci数前40位即可。查询时使用字典树。在计算时，为了保证前40位精确无误。在此我计算了前60位。以保证前面不在进位。

注意点：

       1）关于表示进位问题，需要控制计算的数位数在60以内，切记计算时不要错位（相同位要对齐）。

       2）坑点：题目给出的数插入字典树最好插入前40位即可，否则MLE.

       3）坑点：题目只要求计算下标不超过100000，计算到100000或者以上均会WA，因为输出值不是-1

下面是数组的AC代码：824ms

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<string>
#include<map>
#include<fstream>
#include<ctime>
#include<queue>
#include<vector>
#include<numeric>
#include<string.h>
#include<iomanip>
#include<sstream>
#include<algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef vector<int> BigInterger;
const int maxn = 100000;
const int maxsize = 10;
const int maxnum =65;

struct TrieNode{
	TrieNode(int d = -1) :id(d){ memset(next, NULL, sizeof(next)); }
	int id;
	TrieNode*next[maxsize];
};
TrieNode*T = new TrieNode;

void insert(char*str,const int&index){
	int len = strlen(str);
	TrieNode*p = T;
	int t = 0;
	for (int i = len-1; i>=0&&t<40;i--,t++){
		int id =str[i]-‘0‘;
		if (!p->next[id])p->next[id] = new TrieNode(index);
		p = p->next[id];
	}
}

int search(char*str){
	TrieNode*p = T;
	while (*str != ‘\0‘){
		int id = *str++ - ‘0‘;
		if (!p->next[id])return -1;
		p = p->next[id];
	}
	return p->id;
}
void init(){
	char f[3][maxnum];
	f[1][0] =‘1‘; f[1][1] = ‘\0‘;
	f[0][0] =‘1‘; f[0][1] = ‘\0‘;
	insert("1", 0);
	for (int i = 2; i<maxn; i++){
		int g = 0,j=0,cnt = 0, x;
		int r1 = (i - 1) % 3, r2 = (i - 2) % 3, r = i % 3;
		int lena = strlen(f[r1]);
		int lenb = strlen(f[r2]);
		if (lena >= maxnum - 5){
			memcpy(f[r1], f[r1]+1, lena - 1); f[r1][lena - 1] = ‘\0‘;
			memcpy(f[r2], f[r2]+1, lenb - 1); f[r2][lenb - 1] = ‘\0‘;
			lena--, lenb--;
		}
		while (g || j < lena||j<lenb){
			x = g;
			if (j < lena)x += f[r1][j]-‘0‘;
			if (j < lenb)x += f[r2][j]-‘0‘;
			f[r][cnt++] = x % 10+‘0‘;
			g = x / 10;
			j++;
		}
		f[r][cnt] = ‘\0‘;
		insert(f[r], i);
	}
}
int main(){
	init();
	int T,kase=1;
	char str[50];
	scanf("%d", &T);
	while(T--){
		scanf("%s", str);
		printf("Case #%d: %d\n",kase++,search(str));
	}
	return 0;
}

　采用向量AC的代码624ms(传值要用引用，否则多了不必要的复制)：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<string>
#include<map>
#include<fstream>
#include<ctime>
#include<queue>
#include<vector>
#include<numeric>
#include<string.h>
#include<iomanip>
#include<sstream>
#include<algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef vector<int> BigInterger;
const int maxn = 100000;
const int maxsize = 10;
const int maxnum =65;

struct TrieNode{
	TrieNode(int d = -1) :id(d){ memset(next, NULL, sizeof(next)); }
	int id;
	TrieNode*next[maxsize];
};
TrieNode*T = new TrieNode;

void insert(const BigInterger&v,const int&index){
	int len=v.size();
	TrieNode*p = T;
	int t = 0;
	for (int i = len-1; i>=0&&t<40;i--,t++){
		int id =v[i];
		if (!p->next[id])p->next[id] = new TrieNode(index);
		p = p->next[id];
	}
}

int search(char*str){
	TrieNode*p = T;
	while (*str != ‘\0‘){
		int id = *str++ - ‘0‘;
		if (!p->next[id])return -1;
		p = p->next[id];
	}
	return p->id;
}
//引用传值，避免赋值，效率更高
void add(BigInterger &a, BigInterger &b, BigInterger&c){
	int lena = a.size(), lenb = b.size();
	int g=0, x,i=0;
	c.clear();
	if (lena > maxnum - 5 || lenb > maxnum - 5){ 
		a.erase(a.begin()), b.erase(b.begin());
		lena--, lenb--;
	}
	while (g || i < lena || i < lenb){
		x = g;
		if (i < lena)x += a[i];
		if (i < lenb)x += b[i];
		c.push_back(x % 10);
		g = x / 10;
		i++;
	}
}
void init(){
	BigInterger a[3];
	a[0].push_back(1);
	a[1].push_back(1);
	insert(a[1], 0);
	for (int i = 2; i<maxn; i++){
		add(a[(i - 1) % 3], a[(i - 2) % 3], a[i % 3]);
		insert(a[i % 3], i);
	}
}
int main(){
	init();
	int T,kase=1;
	char str[50];
	scanf("%d", &T);
	while(T--){
		scanf("%s", str);
		printf("Case #%d: %d\n",kase++,search(str));
	}
	return 0;
}