POJ 3982 序列（高精度加法）

数列A满足An = An-1 + An-2 + An-3, n >= 3 

编写程序，给定A0, A1 和 A2, 计算A99

Input

输入包含多行数据 

每行数据包含3个整数A0, A1, A2 (0 <= A0, A1, A2 <= 32767) 
数据以EOF结束

Output

对于输入的每一行输出A99的值

Sample Input

1 1 1

Sample Output

69087442470169316923566147

思路：

直接高精度模拟就好了，直接套模板

代码：

#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>#include<algorithm>using namespace std;#define lson k*2#define rson k*2+1#define M (t[k].l+t[k].r)/2#define INF 100861111#define ll long long#define eps 1e-15class bigInt{public:    int num[302], len;    bigInt() { num[0] = 0, len = 0; }    void operator=(const int &a)    {        int tmp = a;        len = 0;        while (tmp)            num[len++] = tmp % 10, tmp /= 10;        if (!len) num[0] = 0, len = 1;    }    bigInt(const int &a)    {        int tmp = a;        len = 0;        while (tmp)            num[len++] = tmp % 10, tmp /= 10;        if (!len) num[0] = 0, len = 1;    }    bigInt operator+(const bigInt &a)    {        bigInt res;        int i, j, c = 0, adda, addb;        for (i = 0, j = 0; i < len || j < a.len || c; )        {            adda = 0, addb = 0;            if (i < len)                adda = num[i++];            if (j < a.len)                addb = a.num[j++];            res.num[res.len++] = (adda + addb + c) % 10;            c = (adda + addb + c) / 10;        }        return res;    }    void display()    {        int i;        for (i = len - 1; i > 1; i--)            if (num[i]) break;        for (; i >= 0; i--)            printf("%d", num[i]);        printf("\n");    }};int main(){    int a,b,c,i;    while(scanf("%d%d%d",&a,&b,&c)!=EOF)    {        bigInt x,y,z,t;        y=a;        z=b;        t=c;        for(i=1;i<=97;i++)        {            x=y;            y=z;            z=t;            t=x+y;            t=t+z;        }        t.display();    }    return 0;}