cf1207解题报告

cf1207解题报告

A

模拟

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll T,a,b,c,x,y;
int main() {
	cin>>T;
	while(T --> 0) {
		cin>>a>>b>>c>>x>>y;
		ll ans=0;
		if(x>y) {
			while(a>=2&&b>=1) ans+=x,a-=2,b--;
			while(a>=2&&c>=1) ans+=y,a-=2,c--;
		} else {
			while(a>=2&&c>=1) ans+=y,a-=2,c--;
			while(a>=2&&b>=1) ans+=x,a-=2,b--;
		}
		cout<<ans<<"\n";
	}
	return 0;
}

B

能选就选

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int _=110;
int n,m,a[_][_],b[_][_];
vector<pair<int,int> > ans;
int main() {
	cin>>n>>m;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			cin>>a[i][j];
	for(int i=1;i<n;++i) {
		for(int j=1;j<m;++j) {
			if(a[i][j]&&a[i+1][j]&&a[i][j+1]&&a[i+1][j+1]) {
				b[i][j]=b[i+1][j]=b[i][j+1]=b[i+1][j+1]=1;
				ans.push_back(make_pair(i,j));
			}
		}
	}
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			if(a[i][j]!=b[i][j]) return puts("-1"),0;
	printf("%d\n",(int)ans.size());
	for(auto x:ans) printf("%d %d\n",x.first,x.second);
	return 0;
}

C

简单dp

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int _=1e6+7;
ll f[_][2];int s[_];
int main() {
	int T,n,a,b;
	scanf("%d",&T);
	while(T --> 0) {
		scanf("%d%d%d",&n,&a,&b);
		for(int i=1;i<=n;++i) scanf("%1d",&s[i]);
		memset(f,0x3f,sizeof(f));
		f[1][0]=0;
		for(int i=2;i<=n+1;++i) {
			f[i][1]=min(f[i-1][1],f[i-1][0]+a)+b;
			if(!s[i] and !s[i-1])
				f[i][0]=min(f[i-1][0],f[i-1][1]+a);
		}
		ll ans=f[n+1][0]+1LL*n*a+1LL*(n+1)*b;
		cout<<ans<<"\n";
	}
	return 0;
}

D

入门容斥。

\(n!-bad_a-bad_b+bad_a&&bad_b\)

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const ll _=6e5+7,mod=998244353;
ll read() {
	ll x=0,f=1;char s=getchar();
	for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
	for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
	return x*f;
}
ll n,jc[_];
struct node {ll a,b;}c[_];
bool operator == (node x,node y) {
	return x.a==y.a&&x.b==y.b;
}
bool cmp1(node x,node y) {
	return x.a==y.a?x.b<y.b:x.a<y.a;
}
bool cmp2(node x,node y) {
	return x.b==y.b?x.a<y.a:x.b<y.b;
}
int main() {
	n=read();
	jc[0]=jc[1]=1;for(ll i=2;i<=n;++i) jc[i]=jc[i-1]*i%mod;
	for(ll i=1;i<=n;++i) c[i].a=read(),c[i].b=read();
	ll bad1=1,bad2=1;
	sort(c+1,c+1+n,cmp1);
	for(ll l=1,r;l<=n;) {
		r=l;while(c[l].a==c[r+1].a&&r+1<=n) r++;
		bad1*=jc[r-l+1],bad1%=mod;
		l=r+1;
	}
	sort(c+1,c+1+n,cmp2);
	for(ll l=1,r;l<=n;) {
		r=l;while(c[l].b==c[r+1].b&&r+1<=n) r++;
		bad2*=jc[r-l+1],bad2%=mod;
		l=r+1;
	}
	ll ans=jc[n]-bad1-bad2;
	ans=(ans%mod+mod)%mod;
	for(ll i=2;i<=n;++i)
		if(c[i-1].a>c[i].a)
			return cout<<ans<<"\n",0;
	ll good=1;
	for(ll l=1,r;l<=n;) {
		r=l;while(c[l]==c[r+1]&&r+1<=n) r++;
		good*=jc[r-l+1],good%=mod;
		l=r+1;
	}
	ans+=good;
	ans=(ans%mod+mod)%mod;
	cout<<ans<<"\n";
	return 0;
}

E

两次确定x的前7位和后七位。

#include <bits/stdc++.h>
using namespace std;
int tmp,ans;
int main() {
	printf("? ");
	for(int i=1;i<=100;++i) printf("%d ",i);
	printf("\n");
	fflush(stdout);
	scanf("%d",&tmp);
	for(int i=7;i<14;++i) if(tmp&(1<<i)) ans|=1<<i;
	printf("? ");
	for(int i=1;i<=100;++i) printf("%d ",i<<7);
	printf("\n");
	fflush(stdout);
	scanf("%d",&tmp);
	for(int i=0;i<7;++i) if(tmp&(1<<i)) ans|=1<<i;
	cout<<"! "<<ans<<"\n";
	return 0;
}

F

分块。

又读错范围了，开了\(long long T\)飞了.

预处理sum[i][j]\(表示\)%i\(余\)j\(的和。
对于模数大于\)\sqrt{n}$的直接暴力跳。

#include <bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
int read() {
	int x=0,f=1;char s=getchar();
	for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
	for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
	return x*f;
}
const int _=5e5+7;
ll n,sum[1000][1000],a[_];
signed main() {
	n=read();
	int dsr=sqrt(500000);
	while(n --> 0) {
		int opt=read(),x=read(),y=read();
		if(opt==1) {
			a[x]+=y;
			for(int i=1;i<=dsr;++i) sum[i][x%i]+=y;
		} else {//%x=y
			if(x>dsr) {
				int ans=0;
				for(int i=y;i<=500000;i+=x) ans+=a[i];
				cout<<ans<<"\n";
			} else {
				cout<<sum[x][y]<<"\n";
			}
		}
	}
	return 0;
}