题意：给定n道题目，每道题目有k个选项，已知所有正确选项，选对1题得1分
问循环后移一格后总得分s2大于原先总得分s1的方案数
n<=2e5,1<=k<=1e9
思路：特判k=1
easy版本的话写了dp
dp[i][j]表示前i道题循环后s2比s1高j的方案数，j=【-n,n】，n方dp即可
hard版本注意到得分的方案是对称的，即s1>s2和s1<s2的方案数相等
所以问题转化为统计s1=s2的方案数
设s1正确s2不正确的位置为-1,反之为1，其余为0
考虑不同的位置有s个
枚举+1的位置，假设有i个，方案数为C(s,i)
-1的位置同样有i个，方案数为C(s-i,i)
其他s-2*i个位置都是0，每个位置上有(k-2)种选择，方案数为(k-2)^(s-2*i)
剩余相同的位置每个位置上有k种选择，方案数为k^(n-s)
合法的总方案数有k^n种，总方案数-s1=s2的方案数之后除2即为所求
easy版本
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef long double ld; 7 typedef pair<int,int> PII; 8 typedef pair<ll,ll> Pll; 9 typedef vector<int> VI; 10 typedef vector<PII> VII; 11 typedef pair<ll,ll>P; 12 #define N 200000+10 13 #define M 200000+10 14 #define INF 1e9 15 #define fi first 16 #define se second 17 #define MP make_pair 18 #define pb push_back 19 #define pi acos(-1) 20 #define mem(a,b) memset(a,b,sizeof(a)) 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 23 #define lowbit(x) x&(-x) 24 #define Rand (rand()*(1<<16)+rand()) 25 #define id(x) ((x)<=B?(x):m-n/(x)+1) 26 #define ls p<<1 27 #define rs p<<1|1 28 #define fors(i) for(auto i:e[x]) if(i!=p) 29 30 const int MOD=998244353,inv2=(MOD+1)/2; 31 //int p=1e4+7; 32 //double eps=1e-6; 33 int dx[4]={-1,1,0,0}; 34 int dy[4]={0,0,-1,1}; 35 36 ll dp[2010][4100]; 37 int a[N],b[N]; 38 39 int read() 40 { 41 int v=0,f=1; 42 char c=getchar(); 43 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 44 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 45 return v*f; 46 } 47 48 ll readll() 49 { 50 ll v=0,f=1; 51 char c=getchar(); 52 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 53 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 54 return v*f; 55 } 56 57 int main() 58 { 59 //freopen("1.in","r",stdin); 60 //freopen("1.out","w",stdout); 61 int n=read(),k=read(); 62 rep(i,1,n) a[i]=read(); 63 rep(i,1,n) b[i%n+1]=a[i]; 64 if(k==1) 65 { 66 printf("0 "); 67 return 0; 68 } 69 int eps=n; 70 dp[0][0+eps]=1; 71 72 rep(i,1,n) 73 { 74 if(a[i]==b[i]) 75 { 76 rep(j,-n,n) dp[i][j+eps]=dp[i-1][j+eps]*k%MOD; 77 continue; 78 } 79 rep(j,-n,n) dp[i][j+eps]=(dp[i-1][j-1+eps]+dp[i-1][j+1+eps]+1ll*(k-2)*dp[i-1][j+eps]%MOD)%MOD; 80 } 81 ll ans=0; 82 rep(i,1,n) ans=(ans+dp[n][i+eps])%MOD; 83 printf("%I64d ",ans); 84 return 0; 85 }
hard版本
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef long double ld; 7 typedef pair<int,int> PII; 8 typedef pair<ll,ll> Pll; 9 typedef vector<int> VI; 10 typedef vector<PII> VII; 11 typedef pair<ll,ll>P; 12 #define N 200000+10 13 #define M 200000+10 14 #define INF 1e9 15 #define fi first 16 #define se second 17 #define MP make_pair 18 #define pb push_back 19 #define pi acos(-1) 20 #define mem(a,b) memset(a,b,sizeof(a)) 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 23 #define lowbit(x) x&(-x) 24 #define Rand (rand()*(1<<16)+rand()) 25 #define id(x) ((x)<=B?(x):m-n/(x)+1) 26 #define ls p<<1 27 #define rs p<<1|1 28 #define fors(i) for(auto i:e[x]) if(i!=p) 29 30 const int MOD=998244353,inv2=(MOD+1)/2; 31 //int p=1e4+7; 32 //double eps=1e-6; 33 int dx[4]={-1,1,0,0}; 34 int dy[4]={0,0,-1,1}; 35 36 ll fac[N],inv[N]; 37 int a[N],b[N]; 38 39 int read() 40 { 41 int v=0,f=1; 42 char c=getchar(); 43 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 44 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 45 return v*f; 46 } 47 48 ll readll() 49 { 50 ll v=0,f=1; 51 char c=getchar(); 52 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 53 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 54 return v*f; 55 } 56 57 ll C(int x,int y) 58 { 59 if(y>x) return 0; 60 return fac[x]*inv[y]%MOD*inv[x-y]%MOD; 61 } 62 63 ll pw(ll x,ll y) 64 { 65 x%=MOD; 66 ll res=1; 67 while(y) 68 { 69 if(y&1) res=res*x%MOD; 70 x=x*x%MOD; 71 y>>=1; 72 } 73 return res; 74 } 75 76 int main() 77 { 78 //freopen("1.in","r",stdin); 79 //freopen("1.out","w",stdout); 80 int n=read(),k=read(); 81 rep(i,1,n) a[i]=read(); 82 rep(i,1,n) b[i%n+1]=a[i]; 83 if(k==1) 84 { 85 printf("0 "); 86 return 0; 87 } 88 int s=0; 89 rep(i,1,n) 90 if(a[i]!=b[i]) s++; 91 fac[0]=1; 92 rep(i,1,2e5) fac[i]=fac[i-1]*i%MOD; 93 inv[0]=inv[1]=1; 94 rep(i,2,2e5) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD; 95 rep(i,1,2e5) inv[i]=inv[i-1]*inv[i]%MOD; 96 ll ans=0,sum=0; 97 rep(i,0,s/2) 98 { 99 ll tmp=C(s,i)*C(s-i,i)%MOD*pw(k-2,s-i*2)%MOD*pw(k,n-s)%MOD; 100 sum=(sum+tmp)%MOD; 101 } 102 ans=(pw(k,n)-sum+MOD)*inv2%MOD; 103 printf("%I64d ",ans); 104 return 0; 105 }