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 }
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 }
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 }
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 }
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 }
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 }
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 }
时间： 2024-08-28 20:59:14