| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
Re:求助! 2680 Computer Transformation 超时呀!In Reply To:求助! 2680 Computer Transformation 超时呀! Posted by:Silgon at 2009-02-07 23:51:26 > #include <stdio.h>
> #include <string.h>
>
> char f[3][2][400] = {{"0", "1"}, {"0", "1"}, 0};
> char temp[400] = {0}, tmp[400] = {0};
> int n;
> int max(int a, int b)
> {
> if (a > b) return a;
> else return b;
> }
> void fix(char *s)
> {
> int i, len = strlen(s);
> char t;
> for (i = 0; i < len/2; i++)
> {
> t = s[i];
> s[i] = s[len-1-i];
> s[len-1-i] = t;
> }
> }
> char *add(char *s1, char *s2)
> {
> int i, c, len, len1 = strlen(s1), len2 = strlen(s2);
> fix(s1);
> fix(s2);
> len = max(len1, len2);
> for (i = len1; i < len; i++) s1[i] = '0';
> for (i = len2; i < len; i++) s2[i] = '0';
> memset(temp, 0, sizeof(temp));
> for (c = i = 0; i < len; i++)
> {
> c += ((s1[i]-'0')+(s2[i]-'0'));
> temp[i] = (char)(c%10+'0');
> c /= 10;
> }
> if (c > 0) temp[len++] = (char)(c+'0');
> fix(temp);
> fix(s1);
> fix(s2);
> return temp;
> }
>
> int main()
> {
> int i;
> while (scanf("%d", &n) != EOF)
> {
> memset(f, 0, sizeof(f));
> f[0][0][0] = f[1][0][0] = f[2][0][0] = '0';
> f[0][1][0] = f[1][1][0] = f[2][1][0] = '1';
> for (i = 2; i <= n; i++)
> {
> strcpy(f[2][0], add(f[0][0], f[0][1]));
> strcpy(f[2][1], add(f[1][1], strcpy(tmp, f[1][1])));
> strcpy(f[0][0], f[1][0]);
> strcpy(f[0][1], f[1][1]);
> strcpy(f[1][0], f[2][0]);
> strcpy(f[1][1], f[2][1]);
> }
> printf("%s\n", f[2][0]);
> }
> //system("pause");
> return 0;
> }
>
> 试了N次,老是超时!我的递推公式是: f[n]双0的对数 = (f[n-2]双0的对数) + (f[n-2]中1的个数)
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;
}
Followed by: Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator