AtCoder Beginner Contest 172 总结
作者:互联网
颓了三天端午回来打比赛,还是秒切前几道水题但做不动后面的难题,任重道远啊。
前四题思路都比较简单,C枚举一个数组的前缀和并用差值在另外一个数组里二分查找,D可以线性筛约数个数函数然后暴力算,也可以分析每个数的贡献来搞,复杂度都为O(n)。
E没有做出来是真的不爽,只想到了先考虑A,枚举一个排列,A确定以后对B的计算方式确实没有想到。看了题解说的容斥原理恍然大悟。先不考虑A算B有多少种排法(也是),然后减去至少有一位与A相同的排法(),再加回来至少两位与A相同的排法()......
当年搞OI时数学的薄弱板块容斥果然没有放过本人......
F有关博弈论,涉及的Nim游戏还不太会,补了坑再回来写。
A题
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
int main() {
int n=read();
printf("%d\n",n+n*n+n*n*n);
return 0;
}
B题
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
string S,T;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
int main() {
cin>>S>>T;
int n=S.length();
int ans=0;
for (register int i=0;i<n;++i)
ans+=(S[i]!=T[i]);
cout<<ans<<endl;
return 0;
}
C题
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2e5+4;
int n,m;
ll K;
ll a[N],b[N];
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
inline void smax(int &x,int y) {
x=x<y?y:x;
}
int main() {
n=read(),m=read(),K=read();
for (register int i=1;i<=n;++i) a[i]=a[i-1]+read();
for (register int i=1;i<=m;++i) b[i]=b[i-1]+read();
int ans=0;
for (register int i=0;i<=n;++i) {
if (a[i]>K) break;
int res=K-a[i];
int pos=upper_bound(b+1,b+m+1,res)-b-1;
smax(ans,i+pos);
}
printf("%d\n",ans);
return 0;
}
D题
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e7+4;
typedef long long ll;
int n;
int prime[N],tot;
bool vis[N];
int d[N],t[N];
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
inline void linear_shaker(int lim) {
d[1]=1;
for (register int i=2;i<=lim;i++) {
if (!vis[i]) prime[++tot]=i,d[i]=2,t[i]=1;
for (register int j=1;j<=tot&&i*prime[j]<=lim;j++) {
vis[i*prime[j]]=true;
if (i%prime[j]==0) {
t[i*prime[j]]=t[i]+1;
d[i*prime[j]]=d[i]/(t[i]+1)*(t[prime[j]*i]+1);
break;
}
t[i*prime[j]]=1;
d[i*prime[j]]=d[i]*2;
}
}
}
ll ans;
int main() {
n=read();
linear_shaker(n);
for (register int i=1;i<=n;++i)
ans+=1ll*i*d[i];
cout<<ans<<endl;
return 0;
}
E题
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=5e5+5;
const ll MOD=1e9+7;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
ll fac[N]={1,1};
ll inv[N]={1,1};
int n,m;
inline ll fpow(ll a,int b) {
ll ret=1;
while (b) {
if (b&1) ret=ret*a%MOD;
b>>=1,a=a*a%MOD;
}
return ret;
}
inline ll C(int x,int y) {
ll ans=fac[x];
ans=ans*inv[y]%MOD;
ans=ans*inv[x-y]%MOD;
return ans;
}
inline ll P(int x,int y) {
ll ans=fac[x];
ans=ans*inv[x-y]%MOD;
return ans;
}
inline void init() {
for (register int i=1;i<N;++i) fac[i]=fac[i-1]*i%MOD;
inv[N-1]=fpow(fac[N-1],MOD-2);
for (int i=N-2;~i;--i) inv[i]=inv[i+1]*(i+1)%MOD;
}
int main() {
n=read(),m=read();
init();
ll f=P(m,n);
ll g=P(m,n),mk=1;
for (register int i=1;i<=n;++i) {
g-=mk*C(n,i)*P(m-i,n-i);
g=(g%MOD+MOD)%MOD;
mk=-mk;
}
ll ans=f*g%MOD;
cout<<ans<<endl;
return 0;
}
把D题的标准算法也贴一下:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e7+4;
typedef long long ll;
int n;
inline int read() {
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x*f;
}
ll ans;
int main() {
n=read();
for (register int i=1;i<=n;++i) {
int cnt=n/i;
int s=i,t=cnt*i;
ans+=1ll*(s+t)*cnt/2;
}
cout<<ans<<endl;
return 0;
}
标签:AtCoder,Beginner,int,ll,while,ans,include,172,getchar 来源: https://blog.csdn.net/KGV093/article/details/106997614