所以誰告訴我我是越來越傻了呢還是越來越傻了呢還是越來越傻了呢。。。。。。
白金元首與七彩魔法
轉化一堆座標直接算即可。
#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x); i<=(y); i++)
#define per(i,x,y) for (int i=(x); i>=(y); i--)
#define ll long long
#define ld long double
#define inf 1000000000
#define INF 1000000000000000000ll
#define pii pair<int,int>
#define F first
#define S second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define sqr(x) ((x)*(x))
#define cmin(x,y) (x)=(y)<(x)?(y):(x)
#define cmax(x,y) (x)=(y)>(x)?(y):(x)
#define mset(x,y) memset((x),(y),sizeof(x))
#define mcpy(x,y) memcpy((x),(y),sizeof(y))
using namespace std;
const ld pi=acos(-1);
const ld eps=1e-8;
//////////////////////// header files ////////////////////////
ll read(){
char ch=getchar(); ll x=0; int op=1;
for (; !isdigit(ch); ch=getchar()) if (ch=='-') op=-1;
for (; isdigit(ch); ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*op;
}
void write(ll a){
if (a<0) putchar('-'),a=-a;
if (a>=10) write(a/10); putchar(a%10+'0');
}
////////////////////////// fast i/o //////////////////////////
#ifdef mod
ll ksm(ll x,ll p){
ll ret=1;
for (; p; p>>=1,x=x*x%mod) if (p&1) ret=ret*x%mod;
return ret;
}
ll get_inv(ll x){
return ksm(x,mod-2);
}
#else
ll ksm(ll x,ll p){
ll ret=1;
for (; p; p>>=1,x=x*x) if (p&1) ret=ret*x;
return ret;
}
#endif
//////////////////////////// qpow ////////////////////////////
#define y1 y1_
int a1,r1,a2,r2; ld x1,y1,x2,y2,ans;
ld val(ld r,ld g,ld b){
return 0.3*r+0.59*g+0.11*b;
}
ld cal(ld a,ld r){
int h=(int)a/60;
ld f=a/60-h,p=1-r,q=1-f*r,t=1-(1-f)*r;
if (!h) return val(1,t,p);
else if (h==1) return val(q,1,p);
else if (h==2) return val(p,1,t);
else if (h==3) return val(p,q,1);
else if (h==4) return val(t,p,1);
else return val(1,p,q);
}
ld get(ld x,ld y){
ld tmp=atan2(y,x)/pi*180;
tmp=90-tmp; if (tmp<0) tmp+=360;
ld r=sqrt(sqr(x)+sqr(y));
return cal(tmp,r);
}
void solve(){
a1=read(),r1=read(); a2=read(),r2=read();
x1=sin(a1/180.*pi)*r1/100.; y1=cos(a1/180.*pi)*r1/100.;
x2=sin(a2/180.*pi)*r2/100.; y2=cos(a2/180.*pi)*r2/100.;
ans=0;
rep (i,0,10000){//將線段分成1e4份暴力算//1e4就夠了
ld p=i/10000.;
cmax(ans,get(x1*p+x2*(1-p),y1*p+y2*(1-p)));
}
printf("%.4lf\n",(double)ans);
}
#define local
int main(){
#ifdef local
freopen("magika.in","r",stdin); freopen("magika.out","w",stdout);
#endif
int T=read(); while (T--) solve();
return 0;
}
// sample data:
/*
6
30 30 30 30
120 60 120 60
270 100 270 100
30 30 120 60
120 60 270 100
270 100 30 30
0.8785
0.7540
0.2600
0.9704
0.9408
0.8785
*/
// rest:
/*
轉化一堆座標直接算即可...
*/
/*
If you find any bug in my code or solution, please tell me. Thanks so much.
email: [email protected]
*/組合數問題 2
用堆維護，從中間向兩邊擴展。注意組合數比較時，，可以求對數然後比較。
#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x); i<=(y); i++)
#define per(i,x,y) for (int i=(x); i>=(y); i--)
#define ll long long
#define ld long double
#define inf 1000000000
#define INF 1000000000000000000ll
#define pii pair<int,int>
#define F first
#define S second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define sqr(x) ((x)*(x))
#define cmin(x,y) (x)=(y)<(x)?(y):(x)
#define cmax(x,y) (x)=(y)>(x)?(y):(x)
#define mset(x,y) memset((x),(y),sizeof(x))
#define mcpy(x,y) memset((x),(y),sizeof(y))
using namespace std;
const ld pi=acos(-1);
const ld eps=1e-8;
//////////////////////// header files ////////////////////////
ll read(){
char ch=getchar(); ll x=0; int op=1;
for (; !isdigit(ch); ch=getchar()) if (ch=='-') op=-1;
for (; isdigit(ch); ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*op;
}
void write(ll a){
if (a<0) putchar('-'),a=-a;
if (a>=10) write(a/10); putchar(a%10+'0');
}
////////////////////////// fast i/o //////////////////////////
#define mod 1000000007
#ifdef mod
ll ksm(ll x,ll p){
ll ret=1;
for (; p; p>>=1,x=x*x%mod) if (p&1) ret=ret*x%mod;
return ret;
}
ll get_inv(ll x){
return ksm(x,mod-2);
}
#else
ll ksm(ll x,ll p){
ll ret=1;
for (; p; p>>=1,x=x*x) if (p&1) ret=ret*x;
return ret;
}
#endif
//////////////////////////// qpow ////////////////////////////
#define N 1000005
int n,k,fac[N],inv[N],tot,ans,nx,ny; ld Fac[N];
void upd(int &x,int y){ x+=y; if (x>=mod) x-=mod; }
void pre(){
fac[0]=Fac[0]=1;
rep (i,1,N-1){
fac[i]=(ll)fac[i-1]*i%mod;
Fac[i]=Fac[i-1]+(ld)log(i);
}
inv[N-1]=ksm(fac[N-1],mod-2);
per (i,N-2,0) inv[i]=(ll)inv[i+1]*(i+1)%mod;
}
int C(int n,int m){
return (ll)fac[n]*inv[m]%mod*inv[n-m]%mod;
}
struct data{
int x,y; data(int a=0,int b=0){ x=a,y=b; }
bool operator < (const data &t) const{
return Fac[x]+Fac[t.y]+Fac[t.x-t.y]<Fac[t.x]+Fac[y]+Fac[x-y];
}
}a[N];
#define local
int main(){
#ifdef local
freopen("problem2.in","r",stdin); freopen("problem2.out","w",stdout);
#endif
pre();
n=read(); k=read();
if (k==1){
cout<<C(n,n/2)<<endl; exit(0);
}
priority_queue<data> qu;
rep (i,0,n){
qu.push(data(i,i/2));
}
while (k){
data now=qu.top(); qu.pop();
if (now.y<0 || now.y>now.x) continue;
upd(ans,C(now.x,now.y)); k--;
if (now.x/2==now.y){
if (now.x&1){
qu.push(data(now.x,now.x/2+1)); continue;
}
}
if (now.y<now.x/2){
qu.push(data(now.x,now.x-now.y));
} else{
qu.push(data(now.x,now.x-now.y-1));
}
}
cout<<ans<<endl;
return 0;
}
// sample data:
/*
2 3
4
*/
// rest:
/*
*/
/*
If you find any bug in my code or solution, please tell me. Thanks so much.
email: [email protected]
*/最短路
直接建邊，異或的貢獻可以拆開來計算，所以只要建log條邊就行了。。
#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x); i<=(y); i++)
#define per(i,x,y) for (int i=(x); i>=(y); i--)
#define ll long long
#define ld long double
#define inf 1000000000
#define INF 1000000000000000000ll
#define pii pair<int,int>
#define F first
#define S second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define sqr(x) ((x)*(x))
#define cmin(x,y) (x)=(y)<(x)?(y):(x)
#define cmax(x,y) (x)=(y)>(x)?(y):(x)
#define mset(x,y) memset((x),(y),sizeof(x))
#define mcpy(x,y) memset((x),(y),sizeof(y))
using namespace std;
const ld pi=acos(-1);
const ld eps=1e-8;
//////////////////////// header files ////////////////////////
ll read(){
char ch=getchar(); ll x=0; int op=1;
for (; !isdigit(ch); ch=getchar()) if (ch=='-') op=-1;
for (; isdigit(ch); ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*op;
}
void write(ll a){
if (a<0) putchar('-'),a=-a;
if (a>=10) write(a/10); putchar(a%10+'0');
}
////////////////////////// fast i/o //////////////////////////
#ifdef mod
ll ksm(ll x,ll p){
ll ret=1;
for (; p; p>>=1,x=x*x%mod) if (p&1) ret=ret*x%mod;
return ret;
}
ll get_inv(ll x){
return ksm(x,mod-2);
}
#else
ll ksm(ll x,ll p){
ll ret=1;
for (; p; p>>=1,x=x*x) if (p&1) ret=ret*x;
return ret;
}
#endif
//////////////////////////// qpow ////////////////////////////
#define N 1000005
#define M 5000005
int n,m,C,A,B,cnt,head[N],dis[N]; bool vis[N];
struct edge{ int to,nxt,v; }e[M];
void adde(int x,int y,int z){
e[++cnt].to=y; e[cnt].nxt=head[x]; e[cnt].v=z;
head[x]=cnt;
}
struct node{
int dis,x; node(){}
node(int a,int b){ dis=a,x=b; }
bool operator < (const node &t) const{
return dis>t.dis;
}
};
#define local
int main(){
#ifdef local
freopen("path.in","r",stdin); freopen("path.out","w",stdout);
#endif
n=read(); m=read(); C=read();
rep (i,1,m){
int x=read(),y=read(),z=read();
adde(x,y,z);
}
int len=0; for (; n; n>>=1) len++;
A=read(),B=read();
priority_queue<node> q; q.push(node(0,A));
mset(dis,0x3f); dis[A]=0;
while (!q.empty()){
int u=q.top().x; q.pop();
for (int i=head[u],v; i; i=e[i].nxt){
v=e[i].to;
if (dis[v]>dis[u]+e[i].v){
dis[v]=dis[u]+e[i].v;
q.push(node(dis[v],v));
}
}
rep (i,0,len){
int v=u^(1<<i);
if (dis[v]>dis[u]+(1<<i)*C){
dis[v]=dis[u]+(1<<i)*C;
q.push(node(dis[v],v));
}
}
}
cout<<dis[B]<<endl;
return 0;
}
// sample data:
/*
4 2 1
1 3 1
2 4 4
1 4
5
======
7 2 10
1 3 1
2 4 4
3 6
34
*/
// rest:
/*
異或的貢獻是可以分開來計算的，，，
關於異或的邊只需要在每一位上異或即可。。
*/
/*
If you find any bug in my code or solution, please tell me. Thanks so much.
email: [email protected]
*/總結
我好菜啊。
//ps.由於複製了一堆define代碼略顯冗長emm。。
//T4可能要，，以後再補。。