poj 3982- 序列-大整数加法

B - 序列

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3982

Description

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

#include <stdio.h>#include <string.h>#define TT 106int a[TT], b[TT], c[TT];void solve(){int TIME = 99;while (TIME -- ){for (int i=0; i<TT; ++i){a[i] += b[i] + c[i];if (a[i] > 9){ a[i+1] += a[i] / 10;a[i] %= 10;}}for (int i=0; i<TT; ++i){int tmp = a[i];a[i] = b[i];b[i] = c[i];c[i] = tmp;}}TIME = TT;while (!a[TIME--]);for (int i=TIME+1; i>=0; --i)printf("%d", a[i]);printf("\n");}void init(int x, int y, int z){memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));memset(c, 0, sizeof(c));int i=0;while (x > 0){a[i++] = x % 10;x /= 10;}i = 0;while (y > 0){b[i++] = y % 10;y /= 10;}i = 0;while (z > 0){c[i++] = z % 10;z /= 10;}}int main(){int x, y, z;while (~scanf("%d%d%d", &x, &y, &z)){init(x, y, z);solve();}return 0;}