hdu 1715 大菲波数（高精度加法）

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题意求斐波那契数列的第n项。

longlong装不下。

代码：

#include <stdio.h>int a[1005][220];void Fibonacci(){    a[1][0] = 1;    a[1][1] = 1;    a[2][0] = 1;    a[2][1] = 1;    int len = 1;    for (int i = 3; i < 1005; i++)    {        int yu = 0;        for (int j = 1; j <= len; j++)        {            int t = a[i - 1][j] + a[i - 2][j] + yu;            a[i][j] = t % 10;            yu = t / 10;        }        while(yu)        {            a[i][++len] = yu % 10;            yu /= 10;        }        a[i][0] = len;    }}int main(){    Fibonacci();    int ncase;    scanf("%d", &ncase);    while (ncase--)    {        int n;        scanf("%d", &n);        //printf("------%d\n", a[n][0]);        for (int i = a[n][0]; i > 0; i--)//a[n][0] 为数字的长度            printf("%d", a[n][i]);        printf("\n");    }    return 0;}

另一种写法：

#include<iostream>using namespace std;const int N = 1007;int f[N][217];int main(){    int i, j;    f[1][0] = f[2][0] = 1;    for(i = 3; i < N; i++)    {        for(j = 0; j <= 210; j++)            f[i][j] = f[i-1][j]+f[i-2][j];        int flag = 0;        for(j = 0; j <= 210; j++)        {            int tmp = f[i][j] + flag;            flag = tmp/10;            f[i][j] = tmp%10;        }    }    //freopen("data.in", "r", stdin);    int n, pi;    cin >> n;    while(n--)    {        cin >> pi;        i = 210;        while(f[pi][i] == 0)            i--;        cout<<"--------"<<i<<endl;        for(; i >= 0; i--)            cout<<f[pi][i];        cout<<endl;    }    return 0;}

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