2016CCPC杭州

Posted 2020-10-08

A 签到，注意%k

 1 #include <bits/stdc++.h>
 2 const long long mod = 1e9+7;
 3 const double ex = 1e-10;
 4 #define inf 0x3f3f3f3f
 5 using namespace std;
 6 long long a[112345];
 7 int main()
 8 {
 9     int T;
10     cin >> T;
11     for (int cas = 1; cas<=T; cas++){
12         int N;
13         long long K;
14         printf("Case #%d: ",cas);
15         long long sum = 0;
16         scanf("%d%I64d",&N,&K);
17         for (int i = 1; i<=N; i++){
18             scanf("%I64d",&a[i]);
19             sum+=a[i];
20         }
21         if (sum%K){
22             printf("%d\\n",-1);
23             continue;
24         }
25         K = sum/K;
26         sum = 0;
27         long long num = 0;
28         long long ans = 0;
29         for (int i = 1; i<=N;i++){
30             sum+=a[i];
31             if (sum%K==0) {
32                 ans+=((sum/K)-1);
33                 sum=0;
34             } else
35                 ++ans;
36         }
37         printf("%I64d\\n",ans);
38     }
39     return 0;
40 }

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B 有向图强连通

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1005;
 4 int T,V,n,c[maxn],cmp[maxn],boom[maxn],deg[maxn];
 5 vector<int> G[maxn],rG[maxn],vs;
 6 bool used[maxn];
 7 long long x[maxn],y[maxn],r[maxn];
 8 void addedge(int from,int to) {
 9     G[from].push_back(to);
10     rG[to].push_back(from);
11 }
12 void dfs(int v) {
13     used[v]=true;
14     for (int i=0;i<(int)G[v].size();++i)
15         if (!used[G[v][i]])
16             dfs(G[v][i]);
17     vs.push_back(v);
18 }
19 void rdfs(int v,int k) {
20     used[v]=true;
21     cmp[v]=k;
22     for (int i=0;i<(int)rG[v].size();++i)
23         if (!used[rG[v][i]])
24             rdfs(rG[v][i],k);
25 }
26 inline int scc() {
27     memset(used,0,sizeof used);
28     vs.clear();
29     for (int v=0;v<V;++v)
30         if (!used[v])
31             dfs(v);
32     memset(used,0,sizeof used);
33     int k=0;
34     for (int i=vs.size()-1;i>=0;--i)
35         if (!used[vs[i]])
36             rdfs(vs[i],k++);
37     return k;
38 }
39 inline bool check(int i,int j) {
40     if ((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])<=r[i]*r[i])
41         return true;
42     return false;
43 }
44 int main()
45 {
46     scanf("%d",&T);
47     for (int cas=1;cas<=T;++cas) {
48         scanf("%d",&n);
49         V=n;
50         for (int i=0;i<n;++i) {
51             G[i].clear();
52             rG[i].clear();
53         }
54         for (int i=0;i<n;++i)
55             scanf("%I64d%I64d%I64d%d",&x[i],&y[i],&r[i],&c[i]);
56         for (int i=0;i<n;++i)
57             for (int j=0;j<n;++j) if (i!=j)
58                 if (check(i,j))
59                     addedge(i,j);
60         memset(cmp,-1,sizeof cmp);
61         memset(boom,0x3f,sizeof boom);
62         memset(deg,0,sizeof deg);
63         int sz=scc();
64         for (int i=0;i<n;++i)
65             boom[cmp[i]]=min(boom[cmp[i]],c[i]);
66         for (int i=0;i<n;++i)
67             for (int j=0;j<(int)G[i].size();++j) {
68                 int v=G[i][j];
69                 if (cmp[i]!=cmp[v])
70                     ++deg[cmp[v]];
71             }
72         int res=0;
73         for (int i=0;i<sz;++i)
74             if (!deg[i])
75                 res+=boom[i];
76         printf("Case #%d: %d\\n",cas,res);
77     }
78     return 0;
79 }

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C 倒着递推，设最后一次的时间为1

 1 #include <bits/stdc++.h>
 2 const int inf = 0x3f3f3f3f3f3f3f3f;
 3 using namespace std;
 4 const double eps=1e-16;
 5 const int maxn=100005;
 6 int T,n;
 7 long long a[maxn];
 8 int main()
 9 {
10     scanf("%d",&T);
11     for (int cas=1;cas<=T;++cas) {
12         scanf("%d",&n);
13         for (int i=1;i<=n;++i)
14             scanf("%I64d",&a[i]);
15         long long res=0,time=1;;
16         a[n+1] = inf;
17         for (int i=n;i>=1;--i){
18             time = (( a[i] - a[i-1] ) * time + a[i+1]-a[i]-1 ) / (a[i+1] - a[i]);
19             res+=time;
20         }
21         printf("Case #%d: %I64d\\n",cas,res);
22     }
23     return 0;
24 }

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D 拆成两部分，前五位 和 后五位，预处理+查找，map竟然超时，垃圾STL

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int T,K,x,F[10][100001],fact[10][10];
 4 vector<long long> v;
 5 inline int calc(int x,int k) {
 6     int ret=0;
 7     while (x) {
 8         int now=x%10;
 9         ret+=fact[now][k];
10         x/=10LL;
11     }
12     return ret;
13 }
14 int main()
15 {
16     //freopen("in.txt","r",stdin);
17     for (int i=1;i<=9;++i)
18         fact[i][0]=1;
19     for (int i=1;i<=9;++i)
20         for (int j=1;j<=9;++j)
21             fact[i][j]=fact[i][j-1]*i;
22     for (int i=1;i<=9;++i)
23         for (int j=0;j<=100000;++j)
24             F[i][j]=calc(j,i);
25     scanf("%d",&T);
26     for (int cas=1;cas<=T;++cas) {
27         v.clear();
28         scanf("%d%d",&x,&K);
29         for (int i=0;i<100000;++i)
30             v.push_back(i-F[K][i]);
31         sort(v.begin(),v.end());
32         int res=0;
33         for (int i=0;i<=100000;++i) {
34             long long ask=F[K][i]-100000LL*i-x;
35             res+=upper_bound(v.begin(),v.end(),ask)-lower_bound(v.begin(),v.end(),ask);
36         }
37         printf("Case #%d: %d\\n",cas,res-(x==0));
38     }
39     return 0;
40 }

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E 暴力出奇迹

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int T,p[36][3],cnt[10],now[10],res;
 4 inline void change(int x,int v) {
 5     for (int i=0;i<3;++i)
 6         now[p[x][i]]+=v;
 7 }
 8 inline bool check(int x) {
 9     bool flag=true;
10     for (int i=0;i<3;++i) {
11         --now[p[x][i]];
12         if (now[p[x][i]]<0)
13             flag=false;
14     }
15     for (int i=0;i<3;++i)
16         ++now[p[x][i]];
17     return flag;
18 }
19 inline void init() {
20     int x=0;
21     for (int i=1;i<10;++i)
22         for (int j=1;j+i<10;++j) {
23             p[x][0]=i;
24             p[x][1]=j;
25             p[x++][2]=i+j;
26             ++cnt[i];
27             ++cnt[j];
28             ++cnt[i+j];
29         }
30 }
31 void dfs(int u,int v) {
32     res=max(res,v);
33     if (36-u+v<=res||u>36)
34         return;
35     if (check(u)) {
36         change(u,-1);
37         dfs(u+1,v+1);
38         change(u,1);
39     }
40     dfs(u+1,v);
41 }
42 int main()
43 {
44     init();
45     scanf("%d",&T);
46     for (int cas=1;cas<=T;++cas) {
47         for (int i=1;i<=9;++i) {
48             scanf("%d",&now[i]);
49             now[i]=min(now[i],cnt[i]);
50         }
51         res=0;
52         dfs(0,0);
53         printf("Case #%d: %d\\n",cas,res);
54     }
55     return 0;
56 }

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F贪心

 1 #include <bits/stdc++.h>
 2 #define maxn 100010
 3 #define inf 0x3f3f3f3f
 4 #define REP(i,x,y) for(int i=x;i<(y);i++)
 5 #define RREP(i,x,y) for(int i=x;i>(y);i--)
 6 using namespace std;
 7 typedef long long ll;
 8 typedef pair<int,int> pii;
 9 char s[50];
10 int main()
11 {
12     int T,cas=1;scanf("%d",&T);
13     while(T--) {
14         scanf(" %s",s+1);int len=strlen(s+1);
15         ll ans=-inf;
16         for(int pos=2;pos<=len-3;pos++) {
17             ll res1=0;
18             for(int i=2;i<=pos;i++) {
19                 res1=res1*10+(s[i]-‘0‘);
20             }
21             res1+=s[1]-‘0‘;
22             ll res2=0;
23             for(int i=1;i<=pos-1;i++) {
24                 res2=res2*10+(s[i]-‘0‘);
25             }
26             res2+=s[pos]-‘0‘;
27             ll maxx=max(res1,res2);
28            // cout<<res1<<" "<<res2<<" "<<maxx<<endl;
29             ll sub=(s[pos+1]-‘0‘)*(s[pos+2]-‘0‘);
30             ll divide=0;
31             for(int i=pos+3;i<=len;i++) {
32                 divide=divide*10+(s[i]-‘0‘);
33             }
34           //  cout<<divide<<endl;
35           //  cout<<sub<<endl;
36             maxx-=sub/divide;
37             ans=max(ans,maxx);
38         }
39         printf("Case #%d: %lld\\n",cas++,ans);
40     }
41 }

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J 巧妙的问题转化

HDU5942 Just a Math Problem

K 重叠区间直接减去，质数大于2个（区间长度大于777（from 百度数据范围内最远质数间隔）） 其余二分图匹配。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=2005;
 4 int T,n,s,V,match[maxn],sleft,sright,nleft,nright;
 5 vector<int> G[maxn];
 6 bool used[maxn];
 7 inline void addedge(int u,int v) {
 8     G[u].push_back(v);
 9     G[v].push_back(u);
10 }
11 bool dfs(int v) {
12     used[v]=true;
13     for (int i=0;i<(int)G[v].size();++i) {
14         int u=G[v][i],w=match[u];
15         if (w<0||(!used[w]&&dfs(w))) {
16             match[v]=u;
17             match[u]=v;
18             return true;
19         }
20     }
21     return false;
22 }
23 inline int bipartite_matching() {
24     int res=0;
25     memset(match,-1,sizeof match);
26     for (int v=0;v<V;++v)
27         if (match[v]<0) {
28             memset(used,0,sizeof used);
29             if (dfs(v))
30                 ++res;
31         }
32     return res;
33 }
34 int main()
35 {
36     scanf("%d",&T);
37     for (int cas=1;cas<=T;++cas) {
38         scanf("%d%d",&n,&s);
39         sleft=max(n+1,s+1);
40         sright=s+1+n;
41         nleft=1;
42         nright=min(s,n);
43         if (nright-nleft>1000)
44             printf("Case #%d: No\\n",cas);
45         else {
46             V=nright-nleft+1;
47             for (int i=0;i<maxn;++i)
48                 G[i].clear();
49             for (int i=sleft;i<=sright;++i)
50                 for (int j=nleft;j<=nright;++j)
51                     if (i%j==0)
52                         addedge(i-sleft,j-nleft+nright-nleft+1);
53             int sz=bipartite_matching();
54             if (sz==nright-nleft+1)
55                 printf("Case #%d: Yes\\n",cas);
56             else
57                 printf("Case #%d: No\\n",cas);
58         }
59     }
60     return 0;
61 }

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