[ABC195] Panasonic Programming Contest (AtCoder Beginner Contest 195)
Tasks
| Task Name | Time Limit | Memory Limit | ||
|---|---|---|---|---|
| A | Health M Death | 2 sec | 1024 MB | Submit |
| B | Many Oranges | 2 sec | 1024 MB | Submit |
| C | Comma | 2 sec | 1024 MB | Submit |
| D | Shipping Center | 2 sec | 1024 MB | Submit |
| E | Lucky 7 Battle | 2 sec | 1024 MB | Submit |
| F | Coprime Present | 2 sec | 1024 MB | Submit |
A Health M Death
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
int main()
{
// freopen("1.in","r",stdin);
int n,m;
cin>>n>>m;
if(m%n==0) puts("Yes");
else puts("No");
return 0;
}
B Many Oranges
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
LL A,B,W;
int main()
{
// freopen("1.in","r",stdin);
cin>>A>>B>>W;
W*=1000;
LL ans=W/B;
while(ans*B<W) ans++;
if(A*ans>W) puts("UNSATISFIABLE");
else {
printf("%lld ",ans);
ans=W/A;
while((ans+1)*A<=W) ans++;
printf("%lld\n",ans);
}
return 0;
}
C Comma
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
// a num of x bit will use (x-1)/3 ","
LL n,ans;
int main()
{
// freopen("1.in","r",stdin);
cin>>n;
LL bits=1,total=9;
while(n>=total) {
ans+=(bits-1)/3*total;
n-=total;
bits++;
total*=10;
}
ans+=(bits-1)/3*n;
cout<<ans<<endl;
return 0;
}
D Shipping Center
整个人都费用流了。
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int N=512,M=1e6+5;
const LL INF=0x3f3f3f3f3f3f3f3f;
int one[N],idx;
int ver[M],Next[M];
LL edge[M],w[M];
inline void AddEdge(int a,int b,LL c,LL d)
{
Next[idx]=one[a]; ver[idx]=b; edge[idx]=c; w[idx]=d; one[a]=idx++;
Next[idx]=one[b]; ver[idx]=a; edge[idx]=0; w[idx]=-d; one[b]=idx++;
}
int n,m,S,T,Q;
LL dis[N];
int cur[N];
bool vis[N];
queue<int> q;
bool spfa()
{
memset(dis,-0x3f,sizeof dis);
memset(vis,0,sizeof vis);
memcpy(cur,one,sizeof cur);
while(q.size()) q.pop();
int i,x,y;
q.push(S); dis[S]=0;
while(q.size()) {
x=q.front(); q.pop();
vis[x]=false;
for(i=one[x];~i;i=Next[i])
if(edge[i] && dis[y=ver[i]]<dis[x]+w[i]) {
dis[y]=dis[x]+w[i];
if(!vis[y]) q.push(y),vis[y]=true;
}
}
return dis[T]>-INF/2;
}
LL ans;
LL mcmf(int x,LL limit)
{
if(x==T) return limit;
vis[x]=true;
LL k,flow=0;
for(int i=cur[x],y;~i && flow<limit;i=Next[i]) {
y=ver[cur[x]=i];
if(edge[i] && dis[y]==dis[x]+w[i] && !vis[y]) {
k=mcmf(y,min(edge[i],limit-flow));
if(!k) dis[y]=-INF;
edge[i]-=k; edge[i^1]+=k;
flow+=k; ans+=k*w[i];
}
}
vis[x]=false;
return flow;
}
LL ww[N],v[N],siz[N];
void build(int l,int r)
{
memset(one,-1,sizeof one); idx=0;
S=n+m+1,T=n+m+2;
for(int i=1;i<=n;i++) AddEdge(S,i,1,v[i]);
for(int i=1;i<=m;i++)
if(i<l || i>r) AddEdge(i+n,T,1,0);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++) {
if(l<=j && j<=r) continue;
if(ww[i]<=siz[j]) AddEdge(i,j+n,1,0);
}
}
int main()
{
// freopen("1.in","r",stdin);
int i;
int x,y;
scanf("%d%d%d",&n,&m,&Q);
for(i=1;i<=n;i++) scanf("%lld%lld",&ww[i],&v[i]);
for(i=1;i<=m;i++) scanf("%lld",&siz[i]);
while(Q--) {
scanf("%d%d",&x,&y);
build(x,y);
ans=0;
while(spfa())
while((x=mcmf(S,INF)));
// cerr<<x<<endl;
printf("%lld\n",ans);
}
return 0;
}
E Lucky 7 Battle
博弈论，DP。
关于一个“大”数是否是一个“小”数的倍数有个通用的状态设计，就是当前“大”数的前 \(i\) 位 \(\bmod\) “小”数 的余数。
\(z=(10x+y) \bmod 7=((10x \bmod 7)+y) \bmod 7\)
逆推。
\(f[i][j]\) 表示操作到第 i 轮，目前 mod 7 = j , 能否一定 mod 7 余 0 .
若为 Takahashi , 则只要有一个成立即可。
若为 Aoki , 则要全部成立才成立。
Code：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int N=2e5+5;
int n;
char s[N],x[N];
int f[N][7];
// f[i][j] 表示操作到第 i 轮，目前 mod 7 = j , 能否一定 mod 7 余 0 .
// 若为 Takahashi , 则只要有一个成立即可。
// 若为 Aoki , 则要全部成立才成立。
int dp(int i,int j)
{
if(i==n) return (j==0);
if(~f[i][j]) return f[i][j];
if(x[i]=='T') f[i][j]=(dp(i+1,j*10%7)|dp(i+1,(j*10+s[i]-'0')%7));
else f[i][j]=(dp(i+1,j*10%7)&dp(i+1,(j*10+s[i]-'0')%7));
// printf("f[%d][%d] = %d\n",i,j,f[i][j]);
return f[i][j];
}
int main()
{
// freopen("1.in","r",stdin);
memset(f,-1,sizeof f);
scanf("%d",&n);
scanf("%s",s); scanf("%s",x);
if(dp(0,0)) puts("Takahashi");
else puts("Aoki");
return 0;
}
F Coprime Present
还不会。。。
