【POJ】 3982 序列

序列

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7915 Accepted: 3562

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

Source

#include<stdio.h>#include<string.h>const int MAX=100,M=100000000;int a[MAX][100];int b[MAX];void ans(int a0,int a1,int a2){memset(a,0,sizeof(a));a[0][0]=a0,a[1][0]=a1,a[2][0]=a2,b[2]=0;for(int i=3;i<MAX;i++){int k;for(k=0;k<=b[i-1];k++)a[i][k]+=a[i-1][k]+a[i-2][k]+a[i-3][k],a[i][k+1]=a[i][k]/M,a[i][k]%=M;b[i]=k-(a[i][k]?0:1);}int k=b[99];printf("%d",a[99][k]);while(k--&&k>=0)printf("%0.8d",a[99][k]);printf("\n");}int main(){int a0,a1,a2;while(scanf("%d%d%d",&a0,&a1,&a2)!=EOF)ans(a0,a1,a2);return 0;}