AtCoder Beginner Contest 156題解
阿新 • • 發佈:2021-06-23
題目是今天中午我在車站寫的,紀念一下~
前三題很水就不說了。
A
#include<bits/stdc++.h>
using namespace std;
int main(){
int c, r; cin>>c>>r;
r= c>=10? r: (10-c)*100+r;
cout<<r;
return 0;
}
B
#include<bits/stdc++.h> using namespace std; int main(){ int n, b; cin>>n>>b; int cnt=0; while(n) n/=b, cnt++; cout<<cnt; return 0; }
C
#include<bits/stdc++.h> using namespace std; const int N=105; int w[N], n; int main(){ cin>>n; for(int i=1; i<=n; i++) cin>>w[i]; int res=1e9; for(int p=1; p<=100; p++){ int d=0; for(int i=1; i<=n; i++) d+=(w[i]-p)*(w[i]-p); res=min(res, d); } cout<<res; return 0; }
D
容斥。
#pragma GCC optimize("O3") #include<bits/stdc++.h> using namespace std; #define endl '\n' #define debug(x) cerr << #x << ": " << x << endl #define pb(a) push_back(a) #define set0(a) memset(a,0,sizeof(a)) #define rep(i,a,b) for(int i=(a);i<=(b);i++) #define dwn(i,a,b) for(int i=(a);i>=(b);i--) #define ceil(a,b) (a+(b-1))/b #define INF 0x3f3f3f3f #define ll_INF 0x7f7f7f7f7f7f7f7f typedef long long ll; typedef pair<int,int> PII; typedef pair<double,double> PDD; #define int long long inline void read(int &x) { int s=0;x=1; char ch=getchar(); while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();} while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar(); x*=s; } const int N=2e5+5, mod=1e9+7; int n, a, b; int fpow(int x, int p){ int res=1; for(; p; p>>=1, x=x*x%mod) if(p&1) res=res*x%mod; return res; } int fac[N]; void init(){ fac[0]=1; rep(i,1,N-1) fac[i]=fac[i-1]*i%mod; } int inv(int x){ return fpow(x, mod-2); } int C(int a, int b){ int res=1; dwn(i,a,a-b+1) res=res*i%mod; res=res*inv(fac[b])%mod; return res; } signed main(){ init(); read(n), read(a), read(b); int res=((fpow(2, n)-1-C(n, a)-C(n, b))%mod+mod)%mod; cout<<res<<endl; return 0; }
E
組合數知識+容斥
當 \(k\geq n\) 時,我們可以構造出所有情況,轉變為 \(n\) 個不同的盒子放 \(n\) 個相同的球,\(n\) 個盒子都可為空的問題,答案自然是 \(C_{2n-1}^{n-1}\)
否則,必然至少有 \(n-k\) 個盒子非空,我們考慮統計反面:列舉一下 \([1,n-k-1]\) 個盒子為空的貢獻。
#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
#define int long long
inline void read(int &x) {
int s=0;x=1;
char ch=getchar();
while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
x*=s;
}
const int N=4e5+5, mod=1e9+7;
int n, k;
int fpow(int x, int p){
int res=1;
for(; p; p>>=1, x=x*x%mod)
if(p&1) res=res*x%mod;
return res;
}
int fac[N];
void init(){
fac[0]=1;
rep(i,1,N-1) fac[i]=fac[i-1]*i%mod;
}
int inv(int x){
return fpow(x, mod-2);
}
int C(int a, int b){
return fac[a]*inv(fac[a-b])%mod*inv(fac[b])%mod;
}
signed main(){
init();
read(n), read(k);
int res;
res=C(2*n-1, n-1);
if(k<n){ // must be a least n-k bits more than 0
rep(i,1,n-k-1) res=((res-C(n, i)*C(n-1, i-1))%mod+mod)%mod;
}
cout<<res;
return 0;
}
F很怪qwq,會補的