{"id":345,"date":"2013-03-05T21:47:00","date_gmt":"2013-03-05T13:47:00","guid":{"rendered":"http:\/\/nike0good.jp1.rpvhost.net\/249"},"modified":"2013-03-05T21:47:00","modified_gmt":"2013-03-05T13:47:00","slug":"poj_3070_fibonacci-matrix_power","status":"publish","type":"post","link":"https:\/\/nike0good.com\/?p=345","title":{"rendered":"POJ 3070(Fibonacci-\u77e9\u9635\u5e42)"},"content":{"rendered":"\n\n\n
\n
Language:\n\n\n\n
Time Limit:<\/strong> 1000MS<\/td>\n <\/td>\nMemory Limit:<\/strong> 65536K<\/td>\n<\/tr>\n
Total Submissions:<\/strong> 6733<\/td>\n <\/td>\nAccepted:<\/strong> 4765<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Description<\/p>\n

\n<\/span><\/p>\n

 Fibonacci integer sequence\u6307 F<\/em>0<\/sub> = 0, F<\/em>1<\/sub> = 1, and Fn<\/sub><\/em> = Fn<\/sub><\/em> \u2212 1<\/sub> + Fn<\/sub><\/em> \u2212 2<\/sub> for n<\/em> \u2265 2. eg:<\/p>\n

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \u2026<\/p>\n

Fibonacci \u7684\u77e9\u9635\u4e58\u6cd5\u6c42\u6cd5\u5982\u4e0b<\/p>\n

\"\".<\/p>\n

Given an integer n<\/em>, \u6c42 Fn <\/sub><\/em>mod 10000.<\/p>\n<\/div>\n

Input<\/p>\n

\n<\/span><\/p>\n

\u591a\u7ec4\u6570\u636e,\u6bcf\u884c\u4e00\u4e2a n (where 0 \u2264 n<\/em> \u2264 1,000,000,000). \u6570\u636e\u4ee5 \u22121 \u7ed3\u5c3e.<\/p>\n<\/div>\n

Output<\/p>\n

\n<\/span><\/p>\n

\u5bf9\u6bcf\u7ec4\u6570\u636e\u6253\u5370\u4e00\u884c Fn<\/sub><\/em> mod 10000).<\/p>\n<\/div>\n

Sample Input<\/p>\n

0\n9\n999999999\n1000000000\n-1<\/pre>\n
Sample Output<\/p>\n
0\n34\n626\n6875<\/pre>\n
Hint<\/p>\n
\n<\/span><\/p>\n
As a reminder, matrix multiplication is associative, and the product of two 2 \u00d7 2 matrices is given by<\/p>\n
\"\".<\/p>\n
Also, note that raising any 2 \u00d7 2 matrix to the 0th power gives the identity matrix:<\/p>\n
\"\".<\/p>\n<\/div>\n
Source<\/p>\n
\nStanford Local 2006<\/a><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\u6b63\u5b97\u77e9\u9635\u5e42\u7ec3\u624b\u9898<\/span>\u3002<\/span><\/p>\n
\u505a\u4e86\u8fd9\u9898\u5c31\u5dee\u4e0d\u591a\u7406\u89e3\u77e9\u9635\u4e58\u6cd5\u4e86\u3002<\/span><\/p>\n
\u8865\u5145\uff1a\u9012\u5f52\u4e00\u822c\u80fd\u7528\u77e9\u9635\u4e58\u6cd5\uff0c\u5982f[i]=g(f[i-1],f[i-2])...<\/span><\/p>\n
<\/span>    \u4f46\u662f\u89c4\u5212\u4e0d\u884c\uff0c\u5982f[i]=max(....)<\/span><\/p>\n
\n<\/p>\n
<\/span><\/p>\n
#include<cstdio>\n#include<cstring>\n#include<algorithm>\n#include<iostream>\n#include<cstdlib>\n#include<functional>\nusing namespace std;\n#define MAXN (1000000000)\n#define F (10000)\nstruct M\n{\n\tint a[3][3];\n\tM(int i){a[1][1]=a[1][2]=a[2][1]=1;a[2][2]=0;\t}\n\tM(){memset(a,0,sizeof(a));\t}\n\tfriend M operator*(M a,M b)\n\t{\n\t\tM c;\n\t\tfor (int i=1;i<=2;i++)\n\t\t\tfor (int j=1;j<=2;j++)\n\t\t\t\tfor (int k=1;k<=2;k++)\n\t\t\t\t{\n\t\t\t\t\tc.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%F;\n\t\t\t\t}\n\t\treturn c;\n\t}\n\tfriend M pow(M a,int b)\n\t{\n\t\tif (b==1)\n\t\t{\n\t\t\tM c(1);\n\t\t\treturn c;\n\t\t}\n\t\telse\n\t\t{\n\t\t\tM c=pow(a,b\/2);\n\t\t\tc=c*c;\n\t\t\tif (b%2) return c*a;\n\t\t\treturn c;\n\t\t}\n\t}\n};\nint n;\nint main()\n{\n\twhile (cin>>n&&n!=-1)\n\t{\n\t\tif (n==0) printf("0n");\n\t\telse\n\t\t{\n\t\t\tM a=pow(M(1),n);\n\t\t\tprintf("%dn",a.a[1][2]);\n\t\t}\n\t}\n}<\/pre>\n<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
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