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Description
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
Input
The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.
Output
Output the elements of S modulo m in the same way as A is given.
Sample Input
2 2 4
0 1
1 1
Sample Output
1 2
2 3
Source
1 //It is made by jump~ 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <ctime> 9 #include <vector> 10 #include <queue> 11 #include <map> 12 #include <set> 13 using namespace std; 14 typedef long long LL; 15 #define RG register 16 int n,k,MOD; 17 int dui[45],tail; 18 19 struct juz{ 20 LL s[33][33]; 21 }a,c[45],ini,mi[45]; 22 23 inline int getint() 24 { 25 RG int w=0,q=0; char c=getchar(); while((c<‘0‘ || c>‘9‘) && c!=‘-‘) c=getchar(); 26 if (c==‘-‘) q=1, c=getchar(); while (c>=‘0‘ && c<=‘9‘) w=w*10+c-‘0‘, c=getchar(); return q ? -w : w; 27 } 28 29 inline juz jia(juz p,juz q){ 30 juz tmp; 31 for(RG int i=1;i<=n;i++) 32 for(RG int j=1;j<=n;j++) 33 tmp.s[i][j]=p.s[i][j]+q.s[i][j],tmp.s[i][j]%=MOD; 34 return tmp; 35 } 36 37 inline juz cheng(juz p,juz q){ 38 juz tmp; 39 for(RG int i=1;i<=n;i++) for(RG int j=1;j<=n;j++) tmp.s[i][j]=0; 40 for(RG int i=1;i<=n;i++) 41 for(RG int j=1;j<=n;j++) 42 for(RG int l=1;l<=n;l++) 43 tmp.s[i][j]+=p.s[i][l]*q.s[l][j],tmp.s[i][j]%=MOD; 44 return tmp; 45 } 46 47 inline void work(){ 48 n=getint(); k=getint(); MOD=getint(); 49 for(RG int i=1;i<=n;i++) for(RG int j=1;j<=n;j++) ini.s[i][j]=getint(); 50 while(k>0) dui[++tail]=k,k>>=1; mi[tail]=ini; c[tail]=ini; 51 for(RG int i=tail-1;i>=1;i--) { 52 mi[i]=cheng(mi[i+1],mi[i+1]);//每次平方 53 c[i]=jia(c[i+1],cheng(c[i+1],mi[i+1]));//前面的乘以之前的部分再加上自己可降低复杂度 54 if(dui[i]&1) mi[i]=cheng(mi[i],ini),c[i]=jia(c[i],mi[i]); 55 } 56 for(RG int i=1;i<=n;i++) { for(RG int j=1;j<=n;j++) printf("%lld ",c[1].s[i][j]); printf("\n"); } 57 } 58 59 int main() 60 { 61 work(); 62 return 0; 63 }
一个log：
1 //It is made by jump~ 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <ctime> 9 #include <vector> 10 #include <queue> 11 #include <map> 12 #include <set> 13 using namespace std; 14 typedef long long LL; 15 #define RG register 16 int n,k,MOD; 17 int dui[45],tail; 18 19 struct juz{ 20 LL s[33][33]; 21 }a,c[45],ini,mi[45]; 22 23 inline int getint() 24 { 25 RG int w=0,q=0; char c=getchar(); while((c<‘0‘ || c>‘9‘) && c!=‘-‘) c=getchar(); 26 if (c==‘-‘) q=1, c=getchar(); while (c>=‘0‘ && c<=‘9‘) w=w*10+c-‘0‘, c=getchar(); return q ? -w : w; 27 } 28 29 inline juz jia(juz p,juz q){ 30 juz tmp; 31 for(RG int i=1;i<=n;i++) 32 for(RG int j=1;j<=n;j++) 33 tmp.s[i][j]=p.s[i][j]+q.s[i][j],tmp.s[i][j]%=MOD; 34 return tmp; 35 } 36 37 inline juz cheng(juz p,juz q){ 38 juz tmp; 39 for(RG int i=1;i<=n;i++) for(RG int j=1;j<=n;j++) tmp.s[i][j]=0; 40 for(RG int i=1;i<=n;i++) 41 for(RG int j=1;j<=n;j++) 42 for(RG int l=1;l<=n;l++) 43 tmp.s[i][j]+=p.s[i][l]*q.s[l][j],tmp.s[i][j]%=MOD; 44 return tmp; 45 } 46 47 inline void work(){ 48 n=getint(); k=getint(); MOD=getint(); 49 for(RG int i=1;i<=n;i++) for(RG int j=1;j<=n;j++) ini.s[i][j]=getint(); 50 while(k>0) dui[++tail]=k,k>>=1; mi[tail]=ini; c[tail]=ini; 51 for(RG int i=tail-1;i>=1;i--) { 52 mi[i]=cheng(mi[i+1],mi[i+1]);//每次平方 53 c[i]=jia(c[i+1],cheng(c[i+1],mi[i+1]));//前面的乘以之前的部分再加上自己可降低复杂度 54 if(dui[i]&1) mi[i]=cheng(mi[i],ini),c[i]=jia(c[i],mi[i]); 55 } 56 for(RG int i=1;i<=n;i++) { for(RG int j=1;j<=n;j++) printf("%lld ",c[1].s[i][j]); printf("\n"); } 57 } 58 59 int main() 60 { 61 work(); 62 return 0; 63 }
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原文地址：http://www.cnblogs.com/ljh2000-jump/p/5903458.html
