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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9677 Accepted Submission(s):
3210
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int a[10000][260]= {0}; //每个元素可以存储8位数字,所以2005位可以用260个数组元素存储。 5 void init() 6 { 7 int i,j; 8 a[1][0]=1; //赋初值 9 a[2][0]=1; 10 a[3][0]=1; 11 a[4][0]=1; 12 for(i=5; i<10000; i++) 13 { 14 for(j=0; j<260; j++) 15 a[i][j]=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j]; 16 for(j=0; j<260; j++) //每八位考虑进位 17 if(a[i][j]>100000000) 18 { 19 a[i][j+1]+=a[i][j]/100000000; 20 a[i][j]=a[i][j]%100000000; 21 } 22 } 23 } 24 int main() 25 { 26 int n,i,j; 27 init(); 28 while(~scanf("%d",&n)) 29 { 30 for(i=259; i>=0; i--) 31 if(a[n][i]!=0) //不输出高位的0 32 break; 33 printf("%d",a[n][i]); 34 for(j=i-1; j>=0; j--) 35 printf("%08d",a[n][j]); //每个元素存储了八位数字,所以控制输出位数为8,左边补0 36 printf("\n"); 37 } 38 return 0; 39 }
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原文地址:http://www.cnblogs.com/pshw/p/5148787.html