期望：250 100+100+50
实际：210 80+100+30
期望：100 实际：80
最后；两个点T了。可能是求逆元的方法太慢了，也可能是闲的又加了一个快速乘的原因。
#include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define mod 1000000007 using namespace std; int n,m; long long ans; int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } long long qmul(long long x,long long m){ long long ret=0; while(m){ if(m&1) ret=(ret+x)%mod; x=(x+x)%mod; m>>=1; } return ret; } long long ksm(long long a,long long n){ long long ret=1; while(n){ if(n&1) ret=qmul(ret,a)%mod; a=qmul(a,a)%mod; n>>=1; } return ret; } long long ex_gcd(long long a,long long b,long long &x,long long &y,long long &d){ if(!b){ x=1;y=0;d=a; } else{ ex_gcd(b,a%b,y,x,d); y-=(a/b)*x; } } long long inv(long long t,long long p){ long long d,x,y; ex_gcd(t,p,x,y,d); return d==1?(x%p+p)%p:-1; } int main(){ //freopen("lpp.in","r",stdin); freopen("sum.in","r",stdin); freopen("sum.out","w",stdout); n=read();m=read(); for(int i=1;i<=n;i++){ if(i==1){ ans+=m;ans%=mod; } else{ long long tmp=(ksm(i,m)-1)*inv((i-1),mod); ans=ans+qmul(tmp,i); ans%=mod; } } cout<<ans; }
后来明确了，就是因为在快速幂里套了一个快速乘，所以T了两个点。
#include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define mod 1000000007 using namespace std; int n,m; long long ans; int inv[50010]; int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } long long qmul(long long x,long long m){ long long ret=0; while(m){ if(m&1) ret=(ret+x)%mod; x=(x+x)%mod; m>>=1; } return ret; } long long fastpow(long long a,long long b){ long long s=1; for(;b;b>>=1){ if(b&1) s=s*a%mod; a=a*a%mod; } return s; } int main(){ //freopen("lpp.in","r",stdin); freopen("sum.in","r",stdin); freopen("sum.out","w",stdout); n=read();m=read(); inv[1]=1; for(int i=2;i<=50000;i++) inv[i]=(mod-mod/i)*1ll*inv[mod%i]%mod; for(int i=1;i<=n;i++){ if(i==1){ ans+=m;ans%=mod; } else{ ans=ans+((fastpow(i,m)-1)*i)%mod*inv[i-1]; ans%=mod; } } cout<<ans; }
大佬们推公式的思路都好清奇 瑟瑟发抖ing
期望的分：100 实际得分：100
发现大家都用的求最长路，就我思路清奇写了个DP吗？(*^__^*)
/* 期望的分：100 思路：树形dp */ #include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 50010 using namespace std; int n,tot; int dad[MAXN],sum[MAXN],bns[MAXN],size[MAXN],dp[MAXN]; int to[MAXN*2],cap[MAXN*2],net[MAXN*2],head[MAXN]; int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void add(int u,int v,int w){ to[++tot]=v;cap[tot]=w;net[tot]=head[u];head[u]=tot; } void dfs(int now){ size[now]=1; for(int i=head[now];i;i=net[i]) if(dad[now]!=to[i]){ dad[to[i]]=now; dfs(to[i]); sum[now]+=sum[to[i]]+cap[i]; size[now]+=size[to[i]]; } } void pre(int now){ if(size[now]==1) dp[now]=0; for(int i=head[now];i;i=net[i]) if(dad[now]!=to[i]) pre(to[i]); } void work(int now){ int x; for(int i=head[now];i;i=net[i]) if(dad[now]!=to[i]){ work(to[i]); dp[now]=min(dp[to[i]]+cap[i]+(sum[now]-sum[to[i]]-cap[i])*2,dp[now]); } } int main(){ //freopen("lpp.in","r",stdin); freopen("tour.in","r",stdin); freopen("tour.out","w",stdout); n=read(); for(int i=1;i<n;i++){ int u=read(); int v=read(); int w=read(); add(u,v,w);add(v,u,w); } memset(dp,0x7f,sizeof(dp)); dfs(1); pre(1); work(1); cout<<dp[1]; } /* 9 1 2 2 2 4 2 2 3 3 1 5 1 5 6 2 5 7 6 6 8 4 6 9 5 */
期望：50 实际 ：30
读错题目了QwQ
想成了只有4和7是幸运数字。
这样竟然还有30分。。良心的出题人给你 笔芯❤❤❤
/* 期望的分：50 思路：暴力dfs */ #include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define mod 1000000007 using namespace std; int n,k; long long ans; int a[100010],num[100010],vis[100010]; bool vis1[100010]; int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } bool judge(){ memset(vis,0,sizeof(vis)); for(int i=1;i<=k;i++) vis[num[i]]++; for(int i=1;i<=k;i++) if(vis[num[i]]>=2&&vis1[num[i]]) return false; return true; } void dfs(int now,int tot){ if(tot==k){ if(judge()){ ans++; ans%=mod; } return ; } if(now>n) return ; num[tot+1]=a[now];dfs(now+1,tot+1); dfs(now+1,tot); } int main(){ //freopen("lpp.in","r",stdin); freopen("lucky.in","r",stdin); freopen("lucky.out","w",stdout); n=read();k=read(); for(int i=1;i<=n;i++){ a[i]=read(); int x=a[i],fl=0; while(x){ if(x%10!=4&&x%10!=7) fl=1; x/=10; } if(fl==0) vis1[a[i]]=1; else vis1[a[i]]=0; } dfs(1,0); printf("%I64d",ans); }
#include <cstdlib> #include <cstdio> #include <cstring> #include <cmath> #include <iostream> #include <vector> #include <map> using std::min; using std::max; typedef long long i64; const i64 modulo = i64(1e9) + 7; const int MaxN = 100005; const int MaxM = 1024; i64 power(i64 base, i64 k) { i64 res = 1; for ( ; k; k >>= 1) { if (k & 1) res = (res * base) % modulo; base = (base * base) % modulo; } return res; } i64 inv(i64 k) { return power(k, modulo - 2); } std::vector<i64> gen_lucky(i64 upper) { std::vector<i64> res; res.clear(); std::vector<i64> a; a.clear(); a.push_back(0); for (int i = 0; i < 10; ++i) { std::vector<i64> b; b.clear(); for (size_t j = 0; j < a.size(); ++j) { b.push_back(a[j] * 10 + 4); b.push_back(a[j] * 10 + 7); } if (b[0] < upper) { res.insert(res.end(), b.begin(), b.end()); a = b; } else { return res; } } return res; } std::vector<i64> lucky; std::map<i64, int> lucky2idx; int n, m, unlucky, k, a[MaxN], cnt[MaxM]; i64 dp[MaxM][MaxM]; i64 f[MaxN]; i64 bincoef(int a, int b) { i64 tmp = f[a]; tmp = tmp * inv(f[b]) % modulo; tmp = tmp * inv(f[a-b]) % modulo; return tmp; } int main(void) { freopen("lucky.in","r",stdin); freopen("lucky.out","w",stdout); lucky = gen_lucky(1e9); m = lucky.size(); for (int i = 0; i < m; ++i) { lucky2idx[lucky[i]] = i; } scanf("%d%d", &n, &k); unlucky = n; for (int i = 0; i < n; ++i) { scanf("%d", a+i); std::map<i64, int>::iterator it = lucky2idx.find(a[i]); if (it != lucky2idx.end()) ++cnt[it->second], --unlucky; } dp[0][0] = 1; for (int i = 0; i < m; ++i) { for (int j = 0; j <= i; ++j) { dp[i+1][j] = (dp[i+1][j] + dp[i][j]) % modulo; dp[i+1][j+1] += (dp[i+1][j+1] + (dp[i][j] * cnt[i] % modulo)) % modulo; } } f[0] = 1; for (int i = 1; i <= n; ++i) { f[i] = f[i-1] * i % modulo; } i64 ans = 0; int st = max(0, k-m); int ed = min(k, unlucky); for (int i = st; i <= ed; ++i) { ans += dp[m][k-i] * bincoef(unlucky, i) % modulo; ans %= modulo; } printf("%d\n", (int) ans); fclose(stdin); fclose(stdout); return 0; }