Queuing
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2773 Accepted Submission(s): 1275
Problem Description Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time.
Now we define that ‘f’ is shZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcnQgZm9yIGZlbWFsZSBhbmQgoa5toa8gaXMgc2hvcnQgZm9yIG1hbGUuIElmIHRoZSBxdWV1ZaGvcyBsZW5ndGggaXMgTCwgdGhlbiB0aGVyZSBhcmUgMjxzdXA+TDwvc3VwPiBudW1iZXJzIG9mIHF1ZXVlcy4gRm9yIGV4YW1wbGUsIGlmIEwgPSAyLCB0aGVuIHRoZXkgYXJlIGZmLCBtbSwgZm0sIG1mIC4gSWYgdGhlcmUgZXhpc3RzIGEgc3VicXVldWUgYXMgZm1mIG9yIGZmZiwgd2UgY2FsbCBpdCBPLXF1ZXVlCiBlbHNlIGl0IGlzIGEgRS1xdWV1ZS48YnI+CllvdXIgdGFzayBpcyB0byBjYWxjdWxhdGUgdGhlIG51bWJlciBvZiBFLXF1ZXVlcyBtb2QgTSB3aXRoIGxlbmd0aCBMIGJ5IHdyaXRpbmcgYSBwcm9ncmFtLjxicj4KCgogCjxicj4KCklucHV0CgpJbnB1dCBhIGxlbmd0aCBMICgwIDw9IEwgPD0gMTAgPHN1cD42PC9zdXA+KSBhbmQgTS4KCiAKPGJyPgoKT3V0cHV0CgpPdXRwdXQgSyBtb2QgTSgxIDw9IE0gPD0gMzApIHdoZXJlIEsgaXMgdGhlIG51bWJlciBvZiBFLXF1ZXVlcyB3aXRoIGxlbmd0aCBMLgoKIAo8YnI+CgpTYW1wbGUgSW5wdXQKCjxwcmUgY2xhc3M9"brush:java;">3 8 4 7 4 8
Sample Output
6
2
1
记答案为f[n],则易得f[0]=0,f[1]=2,f[2]=4,f[3]=6;f[[4]=9;
当长度为N时，若最后一个字符为M，前N-1个字符没有限制，即为F(N-1);
当最后一个字符串为F的时候，就必须去除最后3个字符是fmf和fff的情况（倒数第二个字符为F、M均有可能会不满足情况），此时最后3个字符可能为mmf和mff；
当后3个字符为mmf时，前N-3个字符没有限制，即F(N-3);
但是当最后四个字符为mmff时，前N-4个字符无限制，即为F(N-1);
即f[n]=f[n-1]+f[n-3]+f[n-4];
转化为矩阵即为：
1 0 1 1 F(N-1) F(N) （即是f[n]=1*f[n-1]+0*f[n-2]+1*f[n-3]+1*f[n-4];）
1 0 0 0 * F(N-2) = F(N-1) （下面为单位矩阵）
0 1 0 0 F(N-3) F(N-2)
0 0 1 0 F(N-4) F(N-3)
#include"iostream"
#include"stdio.h"
#include"string.h"
#include"algorithm"
#include"queue"
#include"vector"
using namespace std;
#define N 4
#define LL __int64
struct Mat
{
LL mat[N][N];
};
int M,n=4;
int p[5]={0,2,4,6,9};
Mat operator *(Mat a,Mat b)
{
int i,j,k;
Mat c;
memset(c.mat,0,sizeof(c.mat));
for(i=0;i
>=1;
a=a*a;
}
LL s=0;
for(i=0;i