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=yif(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;
}