摘要：
解决方案：第一次排序ifbi==bi-1&&ai-ai-1=1返回0；Ifbi==bi-1设f1=3；fn=3^n-fn-1；elsef1＝2；fn=3^n-fn-1；然后使用namespacestd判断两端矩阵-1,3,31#include＜cstio＞2#include＜cs环＞3#include＜iostream＞4#include＜algorithm＞5；6型超长LL；7constlmod=1000000007；8structZ{9LLm[2][2]；10LLa；11stringb；12}s[20]；13ZA＝{14-1,0153316}；17ZB＝{181,0190120}；21LLPow{22LLres=1；23而{24if（b&1）res=res*a%mod；25a=a*a%mod；26b>>=1；27}28返回值；29｝30boolcmp｛31returnp.a＜q.a；32｝33Zoperator*｛34Zc；35 for 36 for｛37c.m[i][j]=0；38 for 39c.m[i][j]=%mod；40｝41returnc；42}43Zmpow{44Zret，p；45ret=B，p=A；46而{47if（n&1）ret=ret*p；48p=p*p；49n>>=1；50}51返回；52｝53intmain（）｛54LLn，m，flag=0；；55while｛56if｛57LLx=4*Pow%mod；58cout>s[i].a>>s[i]。b6364if｛65LLx＝Pow；66cout＜＜x＜＜endl；67继续；68｝69排序；70LLx=1；71Zant；72秒[0]。a=0；73表示{7475if（s[i-1].a！
题目链接：here——————
题意：有四个人 A,B,C,D 每天出一套卷子，相邻的两天不能由同一个人出题
给你两个数n，m分别表示n天和m个操作（把第ai天定为有bi出题）
问有多少种方式？？
题解： 先排序
if bi == bi-1 && ai - ai-1 = 1 return 0;
if bi == bi-1 设f1 = 3；fn = 3^n - fn-1;
else f1 = 2;fn = 3^n - fn-1;
再判断两头
矩阵 -1,0
3,3
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 typedef long long LL; 7 const LL mod = 1000000007; 8 struct Z{ 9 LL m[2][2]; 10 LL a; 11 string b; 12 }s[20]; 13 Z A ={ 14 -1,0, 15 3,3 16 }; 17 Z B = { 18 1,0, 19 0,1 20 }; 21 LL Pow(LL a,LL b){ 22 LL res = 1; 23 while(b){ 24 if(b&1) res = res * a % mod ; 25 a = a * a % mod; 26 b >>= 1; 27 } 28 return res; 29 } 30 bool cmp(Z p,Z q){ 31 return p.a < q.a; 32 } 33 Z operator * (const Z& a,const Z& b){ 34 Z c; 35 for(int i = 0;i < 2 ; ++ i) 36 for(int j = 0;j < 2;j++){ 37 c.m[i][j] = 0; 38 for(int k = 0;k < 2;k++) 39 c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j]) % mod; 40 } 41 return c; 42 } 43 Z mpow(int n){ 44 Z ret , p; 45 ret = B, p = A; 46 while(n){ 47 if(n&1) ret = ret * p; 48 p = p * p; 49 n >>= 1; 50 } 51 return ret; 52 } 53 int main(){ 54 LL n,m,flag = 0;; 55 while(cin>>n>>m){ 56 if(m == 0){ 57 LL x = 4 * Pow(3,n-1) % mod; 58 cout << x << endl; 59 continue; 60 } 61 for(int i = 1;i <= m;i++) 62 cin>>s[i].a>>s[i].b; 63 64 if(m == 1){ 65 LL x = Pow(3,n-1); 66 cout << x << endl; 67 continue; 68 } 69 sort(s+1,s+m+1,cmp); 70 LL x = 1; 71 Z ant; 72 s[0].a = 0; 73 for(int i = 1;i <= m; ++ i){ 74 75 if(s[i-1].a != 0){ 76 LL abs = s[i].a - s[i-1].a - 2; 77 if(abs < 0) 78 { 79 if(s[i-1].b[0] == s[i].b[0]) {x = 0;break;} 80 continue; 81 } 82 ant = mpow(abs); 83 if(s[i-1].b[0] == s[i].b[0]) 84 x = x * (ant.m[0][0]*3 + ant.m[1][0]*3)%mod; 85 else x = x * (ant.m[0][0]*2 + ant.m[1][0]*3)%mod; 86 87 } 88 else { 89 x = x * Pow(3,s[i].a - 1) % mod; 90 } 91 if(i == m && s[i].a < n){ 92 x = x * Pow(3,n-s[i].a) % mod; 93 } 94 } 95 cout << (x%mod + mod)%mod<< endl; 96 } 97 }