2016CCPC杭州

A 签到，注意%k

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
long long a[112345];
int main()
{
    int T;
    cin >> T;
    for (int cas = 1; cas<=T; cas++){
        int N;
        long long K;
        printf("Case #%d: ",cas);
        long long sum = 0;
        scanf("%d%I64d",&N,&K);
        for (int i = 1; i<=N; i++){
            scanf("%I64d",&a[i]);
            sum+=a[i];
        }
        if (sum%K){
            printf("%d\n",-1);
            continue;
        }
        K = sum/K;
        sum = 0;
        long long num = 0;
        long long ans = 0;
        for (int i = 1; i<=N;i++){
            sum+=a[i];
            if (sum%K==0) {
                ans+=((sum/K)-1);
                sum=0;
            } else
                ++ans;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

 

B 有向图强连通

#include <bits/stdc++.h>
using namespace std;
const int maxn=1005;
int T,V,n,c[maxn],cmp[maxn],boom[maxn],deg[maxn];
vector<int> G[maxn],rG[maxn],vs;
bool used[maxn];
long long x[maxn],y[maxn],r[maxn];
void addedge(int from,int to) {
    G[from].push_back(to);
    rG[to].push_back(from);
}
void dfs(int v) {
    used[v]=true;
    for (int i=0;i<(int)G[v].size();++i)
        if (!used[G[v][i]])
            dfs(G[v][i]);
    vs.push_back(v);
}
void rdfs(int v,int k) {
    used[v]=true;
    cmp[v]=k;
    for (int i=0;i<(int)rG[v].size();++i)
        if (!used[rG[v][i]])
            rdfs(rG[v][i],k);
}
inline int scc() {
    memset(used,0,sizeof used);
    vs.clear();
    for (int v=0;v<V;++v)
        if (!used[v])
            dfs(v);
    memset(used,0,sizeof used);
    int k=0;
    for (int i=vs.size()-1;i>=0;--i)
        if (!used[vs[i]])
            rdfs(vs[i],k++);
    return k;
}
inline bool check(int i,int j) {
    if ((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])<=r[i]*r[i])
        return true;
    return false;
}
int main()
{
    scanf("%d",&T);
    for (int cas=1;cas<=T;++cas) {
        scanf("%d",&n);
        V=n;
        for (int i=0;i<n;++i) {
            G[i].clear();
            rG[i].clear();
        }
        for (int i=0;i<n;++i)
            scanf("%I64d%I64d%I64d%d",&x[i],&y[i],&r[i],&c[i]);
        for (int i=0;i<n;++i)
            for (int j=0;j<n;++j) if (i!=j)
                if (check(i,j))
                    addedge(i,j);
        memset(cmp,-1,sizeof cmp);
        memset(boom,0x3f,sizeof boom);
        memset(deg,0,sizeof deg);
        int sz=scc();
        for (int i=0;i<n;++i)
            boom[cmp[i]]=min(boom[cmp[i]],c[i]);
        for (int i=0;i<n;++i)
            for (int j=0;j<(int)G[i].size();++j) {
                int v=G[i][j];
                if (cmp[i]!=cmp[v])
                    ++deg[cmp[v]];
            }
        int res=0;
        for (int i=0;i<sz;++i)
            if (!deg[i])
                res+=boom[i];
        printf("Case #%d: %d\n",cas,res);
    }
    return 0;
}

 

C 倒着递推，设最后一次的时间为1

#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f3f3f3f3f;
using namespace std;
const double eps=1e-16;
const int maxn=100005;
int T,n;
long long a[maxn];
int main()
{
    scanf("%d",&T);
    for (int cas=1;cas<=T;++cas) {
        scanf("%d",&n);
        for (int i=1;i<=n;++i)
            scanf("%I64d",&a[i]);
        long long res=0,time=1;;
        a[n+1] = inf;
        for (int i=n;i>=1;--i){
            time = (( a[i] - a[i-1] ) * time + a[i+1]-a[i]-1 ) / (a[i+1] - a[i]);
            res+=time;
        }
        printf("Case #%d: %I64d\n",cas,res);
    }
    return 0;
}

 

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

#include <bits/stdc++.h>
using namespace std;
int T,K,x,F[10][100001],fact[10][10];
vector<long long> v;
inline int calc(int x,int k) {
    int ret=0;
    while (x) {
        int now=x%10;
        ret+=fact[now][k];
        x/=10LL;
    }
    return ret;
}
int main()
{
    //freopen("in.txt","r",stdin);
    for (int i=1;i<=9;++i)
        fact[i][0]=1;
    for (int i=1;i<=9;++i)
        for (int j=1;j<=9;++j)
            fact[i][j]=fact[i][j-1]*i;
    for (int i=1;i<=9;++i)
        for (int j=0;j<=100000;++j)
            F[i][j]=calc(j,i);
    scanf("%d",&T);
    for (int cas=1;cas<=T;++cas) {
        v.clear();
        scanf("%d%d",&x,&K);
        for (int i=0;i<100000;++i)
            v.push_back(i-F[K][i]);
        sort(v.begin(),v.end());
        int res=0;
        for (int i=0;i<=100000;++i) {
            long long ask=F[K][i]-100000LL*i-x;
            res+=upper_bound(v.begin(),v.end(),ask)-lower_bound(v.begin(),v.end(),ask);
        }
        printf("Case #%d: %d\n",cas,res-(x==0));
    }
    return 0;
}

 

E 暴力出奇迹

#include <bits/stdc++.h>
using namespace std;
int T,p[36][3],cnt[10],now[10],res;
inline void change(int x,int v) {
    for (int i=0;i<3;++i)
        now[p[x][i]]+=v;
}
inline bool check(int x) {
    bool flag=true;
    for (int i=0;i<3;++i) {
        --now[p[x][i]];
        if (now[p[x][i]]<0)
            flag=false;
    }
    for (int i=0;i<3;++i)
        ++now[p[x][i]];
    return flag;
}
inline void init() {
    int x=0;
    for (int i=1;i<10;++i)
        for (int j=1;j+i<10;++j) {
            p[x][0]=i;
            p[x][1]=j;
            p[x++][2]=i+j;
            ++cnt[i];
            ++cnt[j];
            ++cnt[i+j];
        }
}
void dfs(int u,int v) {
    res=max(res,v);
    if (36-u+v<=res||u>36)
        return;
    if (check(u)) {
        change(u,-1);
        dfs(u+1,v+1);
        change(u,1);
    }
    dfs(u+1,v);
}
int main()
{
    init();
    scanf("%d",&T);
    for (int cas=1;cas<=T;++cas) {
        for (int i=1;i<=9;++i) {
            scanf("%d",&now[i]);
            now[i]=min(now[i],cnt[i]);
        }
        res=0;
        dfs(0,0);
        printf("Case #%d: %d\n",cas,res);
    }
    return 0;
}

 

F贪心

#include <bits/stdc++.h>
#define maxn 100010
#define inf 0x3f3f3f3f
#define REP(i,x,y) for(int i=x;i<(y);i++)
#define RREP(i,x,y) for(int i=x;i>(y);i--)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
char s[50];
int main()
{
    int T,cas=1;scanf("%d",&T);
    while(T--) {
        scanf(" %s",s+1);int len=strlen(s+1);
        ll ans=-inf;
        for(int pos=2;pos<=len-3;pos++) {
            ll res1=0;
            for(int i=2;i<=pos;i++) {
                res1=res1*10+(s[i]-'0');
            }
            res1+=s[1]-'0';
            ll res2=0;
            for(int i=1;i<=pos-1;i++) {
                res2=res2*10+(s[i]-'0');
            }
            res2+=s[pos]-'0';
            ll maxx=max(res1,res2);
           // cout<<res1<<" "<<res2<<" "<<maxx<<endl;
            ll sub=(s[pos+1]-'0')*(s[pos+2]-'0');
            ll divide=0;
            for(int i=pos+3;i<=len;i++) {
                divide=divide*10+(s[i]-'0');
            }
          //  cout<<divide<<endl;
          //  cout<<sub<<endl;
            maxx-=sub/divide;
            ans=max(ans,maxx);
        }
        printf("Case #%d: %lld\n",cas++,ans);
    }
}

 

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。

#include <bits/stdc++.h>
using namespace std;
const int maxn=2005;
int T,n,s,V,match[maxn],sleft,sright,nleft,nright;
vector<int> G[maxn];
bool used[maxn];
inline void addedge(int u,int v) {
    G[u].push_back(v);
    G[v].push_back(u);
}
bool dfs(int v) {
    used[v]=true;
    for (int i=0;i<(int)G[v].size();++i) {
        int u=G[v][i],w=match[u];
        if (w<0||(!used[w]&&dfs(w))) {
            match[v]=u;
            match[u]=v;
            return true;
        }
    }
    return false;
}
inline int bipartite_matching() {
    int res=0;
    memset(match,-1,sizeof match);
    for (int v=0;v<V;++v)
        if (match[v]<0) {
            memset(used,0,sizeof used);
            if (dfs(v))
                ++res;
        }
    return res;
}
int main()
{
    scanf("%d",&T);
    for (int cas=1;cas<=T;++cas) {
        scanf("%d%d",&n,&s);
        sleft=max(n+1,s+1);
        sright=s+1+n;
        nleft=1;
        nright=min(s,n);
        if (nright-nleft>1000)
            printf("Case #%d: No\n",cas);
        else {
            V=nright-nleft+1;
            for (int i=0;i<maxn;++i)
                G[i].clear();
            for (int i=sleft;i<=sright;++i)
                for (int j=nleft;j<=nright;++j)
                    if (i%j==0)
                        addedge(i-sleft,j-nleft+nright-nleft+1);
            int sz=bipartite_matching();
            if (sz==nright-nleft+1)
                printf("Case #%d: Yes\n",cas);
            else
                printf("Case #%d: No\n",cas);
        }
    }
    return 0;
}