Questions surface: https://www.cnblogs.com/Juve/articles/11790223.html
96：
Thought it was just a look at water problems, the number of direct arithmetic sequence summation just fine, and then found the modulus is not a prime number, but also 1e18 * 1e18, it abandoned, and look at T3, then the wrong question, called dij 40 sub-violence
Then I feel like I'll see T1 find a ride called the slow stuff (to the decadent decadent AlpaCa blog, it is now my second template), then do not play a high-precision,
As for analog to digital is not a prime number, because the answer must be an integer, and my final formula is divided by 4, so I let it take before the addition and multiplication, and then T1 to A, and
N T2 called party violence, lured 75, rk6 still the best in recent times? I may be too much garbage. . .
T1:
Above have said, not elaborate
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define int long long 6 using namespace std; 7 int x,y,xx,yy,tot=0; 8 int ans=0,mod; 9 int mul(int a,int b,int p){ 10 int res=0; 11 while(b){ 12 if(b&1) res=(res+a)%p; 13 a=(a+a)%p; 14 b>>=1; 15 } 16 return res; 17 } 18 signed main(){ 19 freopen("sum.in","r",stdin); 20 freopen("sum.out","w",stdout); 21 scanf("%lld%lld%lld%lld%lld",&x,&y,&xx,&yy,&mod); 22 int p=(x+y-1),q=(x+yy-1),pp=(xx+y-1),qq=(xx+yy-1); 23 int xkl1=(p+q+pp+qq),xkl2=(yy-y+1),xkl3=(xx-x+1); 24 while(tot<2&&xkl1%2==0){ 25 xkl1/=2; 26 ++tot; 27 } 28 while(tot<2&&xkl2%2==0){ 29 xkl2/=2; 30 ++tot; 31 } 32 while (tot < 2 && xkl3% 2 == 0 ) { 33 xkl3 / = 2 ; 34 ++ tot; 35 } 36 ans = mul (mul (xkl2, xkl3, v) v%, xkl1, v)% v; 37 printf ( " % lld \ n " , ANS); 38 return 0 ; 39 }
T2:
Just greedy sweep, half the maximum feasible location, but the n- 2 logN 2 complexity are not good enough, we doubled narrow-half interval
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<vector> 6 #define int long long 7 #define re register 8 using namespace std; 9 inline int read(){ 10 re int x=0;re char ch=getchar(); 11 while(ch<'0'||ch>'9') ch=getchar(); 12 while(ch>='0'&&ch<='9'){ 13 x=(x<<3)+(x<<1)+ch-'0'; 14 ch=getchar(); 15 } 16 return x; 17 } 18 const int MAXN=1e6+5; 19 int n,m,a[MAXN],b[MAXN],ans=0; 20 int staa[MAXN],stab[MAXN],topa,topb; 21 bool check(int l,int r){ 22 topa=topb=0; 23 for(int i=l;i<=r;++i) staa[++topa]=a[i],stab[++topb]=b[i]; 24 sort(staa+1,staa+topa+1),sort(stab+1,stab+topb+1); 25 int tot=0; 26 for(int i=1;i<=topa;++i){ 27 tot+=staa[i]*stab[i]; 28 if(tot>m) return 0; 29 } 30 return 1; 31 } 32 int get(int pos){ 33 int poss=1; 34 for(int i=1;;++i){ 35 poss=i; 36 if(pos+(1<<i)-1>n) break; 37 if(!check(pos,pos+(1<<i)-1)){ 38 poss=i; 39 break; 40 } 41 } 42 int l=pos+(1<<(poss-1))-1,r=pos+(1<<poss)-1; 43 int res=l; 44 while(l<=r){ 45 int mid=(l+r)>>1; 46 if(check(pos,mid)) res=max(res,mid),l=mid+1; 47 else r=mid-1; 48 } 49 return res+1; 50 } 51 signed main(){ 52 freopen("pair.in","r",stdin); 53 freopen("pair.out","w",stdout); 54 n=read(),m=read(); 55 for(re int i=1;i<=n;++i) a[i]=read(); 56 for(re int i=1;i<=n;++i) b[i]=read(); 57 for(re int i=1;i<=n;){ 58 i=get(i); 59 ++ans; 60 } 61 printf("%lld\n",ans); 62 return 0; 63 }
T3:
Less likely
97：
T1 discovered the nature ran 50 minutes, T2 is a feeling of sb dp, but because I set the state causes him to have eight transfers, complexity only had 70 points, T3 metaphysical reasons for the violence are linked to 10 points
dp again review, dp slag what to say?
Solution to a problem I do not find that issued the interpretation of T1, T2 and too much water, causing nothing to speak of
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define int long long 6 using namespace std; 7 const int MAXN=1e5+5,mod=1e9+7; 8 int n,a[MAXN],cnt=0,ans=0,tong[MAXN]; 9 int q_pow(int a,int b,int p){ 10 int res=1; 11 while(b){ 12 if(b&1) res=res*a%p; 13 a=a*a%p; 14 b>>=1; 15 } 16 return res; 17 } 18 signed main(){ 19 freopen("game.in","r",stdin); 20 freopen("game.out","w",stdout); 21 scanf("%lld",&n); 22 for(int i=1;i<=n;++i){ 23 scanf("%lld",&a[i]); 24 ++tong[a[i]]; 25 cnt+=(a[i]==-1); 26 } 27 ans=(q_pow(2,n-1,mod)-1+mod)%mod; 28 for(int i=1;i<=n;++i){ 29 ans=(ans-q_pow(2,tong[i],mod)+1+mod)%mod; 30 } 31 printf("%lld\n",ans); 32 return 0; 33 }
T2:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define int long long 6 #define re register 7 using namespace std; 8 const int mod=1e9+7; 9 int t,n,s,f[100005][2]; 10 signed main(){ 11 freopen("flower.in","r",stdin); 12 freopen("flower.out","w",stdout); 13 scanf("%lld",&t); 14 while(t--){ 15 scanf("%lld%lld",&n,&s); 16 f[1][0]=f[3][1]=s; 17 f[2][0]=s*s%mod; 18 f[3][0]=(s*s%mod*s%mod-s+mod)%mod; 19 for(int i=4;i<=n;++i){ 20 f[i][0]=(f[i-1][0]*(s-1)%mod+f[i-2][0]*(s-1)%mod)%mod; 21 f[i][1]=(f[i-1][1]*(s-1)%mod+f[i-2][1]*(s-1)%mod+f[i-3][0]*(s-1)%mod)%mod; 22 } 23 printf("%lld\n",f[n][1]%mod); 24 } 25 return 0; 26 }
T3:
Not, DEE titanium saw wax tree
98：
Full stroke, then T130 sub violence hung up, because the loop limit the number of layers is too small, did not run out of leads
T2 form a pressure error dp water 35 minutes
Well, I now only T2
01 set state, the pretreatment each point i, the lamp lighting state that it will be negated j seconds, then backwards transferred
Saying Yu-shi told me when I have been interpreted as positive sequence also convinced myself that I really bang up
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int n,fa[20],zt[20],s,dp[17][17],ans=0x3f3f3f3f; 7 bool f[17][(1<<16)+5]; 8 int calc(int state){ 9 int res=0; 10 for(int i=2;i<=n;++i){ 11 if(state&(1<<(i-1))){ 12 res^=(1<<(fa[i]-1)); 13 } 14 } 15 return res; 16 } 17 void print(int sta){ 18 for(int i=1;i<=n;++i){ 19 if(sta&(1<<(i-1))) cout<<1; 20 else cout<<0; 21 } 22 cout<<' '; 23 for(int i=n;i>=1;--i){ 24 if(sta&(1<<(i-1))) cout<<1; 25 else cout<<0; 26 } 27 cout<<' '; 28 } 29 int main(){ 30 freopen("decoration.in","r",stdin); 31 freopen("decoration.out","w",stdout); 32 scanf("%d",&n); 33 for(int i=2;i<=n;++i) scanf("%d",&fa[i]); 34 for(int i=1;i<=n;++i){ 35 scanf("%d",&zt[i]); 36 s|=(zt[i]<<(i-1)); 37 int p=i; 38 dp[i][1]=1<<(i-1); 39 p=fa[p]; 40 for(int j=2;j<=n;++j,p=fa[p]){ 41 if(p!=0) dp[i][j]=dp[i][j-1]|(1<<(p-1)); 42 else dp[i][j]=dp[i][j-1]; 43 } 44 } 45 f[0][0]=1; 46 for(int i=1;i<=n;++i){ 47 for(int s=0;s<(1<<n);++s){ 48 f[i][s]|=f[i-1][s]; 49 for(int j=1;j<=n;++j){ 50 f[i][s^dp[j][i]]|=f[i-1][s]; 51 } 52 } 53 } 54 for(int i=0;i<=n;++i){ 55 if(f[i][s]){ 56 ans=i; 57 break; 58 } 59 } 60 printf("%d\n",ans); 61 return 0; 62 }