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## Re:规律

Posted by Silgon at 2009-02-08 13:13:36 on Problem 2680
In Reply To:Re:规律 Posted by:Silgon at 2009-02-08 00:01:04
```>
> 可以这样吗？
> 用f[n][0]表示n中00的对数，用f[n][1]表示n中1的个数
> 当n＝0时，f[n][0] = 0, f[n][1] = 1;
> 当n=1时， f[n][0] = 0, f[n][1] = 1;
> 当n>=2时，f[n][0] = f[n-2][0]+f[n-2][1];
>          f[n][1] = 2 * f[n][1];
> 则f[n][0]为所求解。
AC了！！！
#include <stdio.h>

int result[1001][401], n;

void compute(int a[], int c[], int n)
{
int i, temp;
memset(c, 0, sizeof(result[0]));
if (n % 2 == 0)    c[1]++;
else    c[1]--;
for (i=1; i<=400; i++) {
temp = a[i] + a[i] + c[i];
if(temp < 0) {
temp += 10;
c[i] = temp % 10;
c[i+1]--;
continue;
}
c[i] = temp % 10;
c[i+1] = temp / 10;
}
}

int main()
{
int i, j;
memset(result, 0, sizeof(result));
for(i = 2; i<=1000; i++)         compute(result[i-1], result[i], i);
while (scanf("%d", &n) != EOF) {
if(n == 0) {
printf("0\n");
continue;
}
if(n == 1) {
printf("0\n");
continue;
}
for(i=400; i>0; i--)
if(result[n][i] != 0)   {
j = i;
break;
}
for (; j > 0; j--)
printf("%d", result[n][j]);
printf("\n");
}
//system("pause");
return 0;
}
```

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