哈尔滨理工大学软件与微电子学院第八届程序设计竞赛同步赛（高年级） Solution

A：

Solved.

分别处理出每个%7后余数的数字个数，再组合一下

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 int n, m;
 ll cnt[][];

 ll calc(ll x, ll y)
 {
     if (x %  >= y) return x /  + ;
     return x / ;
 }

 int main()
 {
     while (scanf("%d%d", &n, &m) != EOF)
     {
         memset(cnt, , sizeof cnt);
         for (int i = ; i < ; ++i)
         {
             cnt[][i] += calc(n, i);
             cnt[][i] += calc(m, i);
         }
         ll res = ;
         --cnt[][], --cnt[][];
         cnt[][] = cnt[][];
         for (int i = ; i < ; ++i)
             res += cnt[][i] * cnt[][ - i];
         printf("%lld\n", res);
     }
     return ;
 }

B：

Solved.

状压DP.

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 int n,m;
 ll f[][<<];
 bool ins[<<];
 int main(){
         while(cin>>n>>m&&n){
                 for(int i=;i<(<<m);i++){
                         bool cnt=,hasodd=;
                         for(int j=;j<m;j++)
                                 if((i>>j)&)hasodd|=cnt,cnt=;
                                 else  cnt^=;
                         ins[i]=hasodd|cnt?:;

                 }
                 f[][]=;
                 for(int i=;i<=n;i++)
                         for(int j=;j<(<<m);j++){
                                 f[i][j]=;
                                 for(int k=;k<(<<m);k++){
                                         if((j&k)==&&ins[j|k])
                                                 f[i][j]+=f[i-][k];
                                 }

                         }
                 cout<<f[n][]<<endl;
         }
 }

C：

Solved.

 #include<bits/stdc++.h>

 using namespace std;

 int n, s;

 int main()
 {
     while(~scanf("%d %d", &n, &s))
     {
         int Min = 0x3f3f3f3f;
         for(int i = ; i < n; ++i)
         {
             int num = ;
             scanf("%d" , &num);
             Min = min(Min, num);
         }
         printf("%d\n", s * Min);
     }
     return ;
 }

D：

Solved.

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn = ;
 int n, s;
 int arr[maxn];

 int main()
 {
     while(~scanf("%d %d", &n, &s))
     {
         for(int i = ; i <= n; ++i)
         {
             scanf("%d", arr + i);
         }
         sort(arr + , arr +  + n);
         printf("%d\n", arr[n - ] * s);
     }
     return ;
 }

E：

Solved.

最长公共子序列。

 #include<bits/stdc++.h>
 using namespace std;
 const int maxn=;
 int f[maxn][maxn];
 char s1[maxn],s2[maxn];
 int main(){
     cin>>s1+>>s2+;
     int n=strlen(s1+),m=strlen(s2+);
     for(int i=;i<=n;i++)
     {

             for(int j=;j<=m;j++)
             {
                     f[i][j]=max(f[i - ][j], f[i][j - ]);
                     if(s1[i]==s2[j])
                     {
                         f[i][j]=max(f[i][j],f[i-][j-] + );
                     }

             }
     }
     printf("%d\n",f[n][m]);
 }

F：

Solved.

$dp[i][j][k] 表示到达i, j 的时候步数为k的方案数$

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn=;
 const ll MOD = 1e9 + ;
 ll dp[maxn][maxn][maxn];
 int fx[][]={{,},{,-},{-,},{-,-},{,},{,-},{-,},{-,-}};
 int n,m,s;
 inline bool  check(int x,int y){
         if(x< || x>n || y< ||y>m)return false;
         return true;
 }
 int main(){
     while(cin>>n>>m>>s)
     {
             memset(dp, , sizeof dp);
         dp[][][]=;
         for(int k=;k<=s;k++)
         {
             for(int i=;i<=n;i++)
             {
                 for(int j=;j<=m;j++){
                     for(int t=;t<;t++)
                     {
                         int x=i+fx[t][];
                         int y=j+fx[t][];
                         if(check(x,y)){
                             dp[i][j][k]+=dp[x][y][k-];
                             dp[i][j][k]%=MOD;
                         }
                     }
                 }
             }
         }
         cout<<dp[n][m][s]<<endl;
     }
 }

G：

Unsolved.

H：

Solved.

考虑离线，对$每个数处理出到L左边离它最近的非互质的数，以及到R右边$离它最近的非互质的数，

再考虑先固定$l, 对每一个r处理出答案$

再考虑什么时候把数的贡献加进去，仅当这个数的$L < l 的时候便可以加入贡献，贡献的范围是[pos, R - 1], pos 表示那个数的位置$

 #include <bits/stdc++.h>
 using namespace std;

 #define N 100010
 #define pii pair <int, int>
 int n, q, a[N], l[N], r[N], dp[N], ans[N], vis[N];
 vector <int> fac[N];
 vector <pii> v[N], qv[N];

 namespace SEG
 {
     int a[N << ], lazy[N << ];
     void build(int id, int l, int r)
     {
         a[id] = ;
         if (l == r) return;
         int mid = (l + r) >> ;
         build(id << , l, mid);
         build(id <<  | , mid + , r);
     }
     void pushdown(int id, int l, int r, int mid)
     {
         if (!lazy[id]) return;
         lazy[id << ] += lazy[id];
         a[id << ] += lazy[id] * (mid - l + );
         lazy[id <<  | ] += lazy[id];
         a[id <<  | ] += lazy[id] * (r - mid);
         lazy[id] = ;
     }
     void pushup(int id) { a[id] = a[id << ] + a[id <<  | ]; }
     void update(int id, int l, int r, int ql, int qr, int val)
     {
         if (l >= ql && r <= qr)
         {
             a[id] += val * (r - l + );
             lazy[id] += val;
             return;
         }
         int mid = (l + r) >> ;
         pushdown(id, l, r, mid);
         if (ql <= mid) update(id << , l, mid, ql, qr, val);
         if (qr > mid) update(id <<  | , mid + , r, ql, qr, val);
         pushup(id);
     }
     int query(int id, int l, int r, int pos)
     {
         if (l == r) return a[id];
         int mid = (l + r) >> ;
         pushdown(id, l, r, mid);
         if (pos <= mid) return query(id << , l, mid, pos);
         else return query(id <<  | , mid + , r, pos);
     }
 }

 void init()
 {
     for (int i = ; i <= ; ++i)
     {
         if (vis[i]) continue;
         fac[i].push_back(i);
         for (int j =  * i; j <= ; j += i)
         {
             vis[j] = ;
             fac[j].push_back(i);
         }
     }
 }

 int main()
 {
     init();
     while (scanf("%d%d", &n, &q) != EOF)
     {
         for (int i = ; i <= n + ; ++i) l[i] = , r[i] = n + , v[i].clear(), qv[i].clear();
         for (int i = ; i <= n; ++i) scanf("%d", a + i);
         memset(vis, , sizeof vis);
         for (int i = ; i <= n; ++i)
             for (auto it : fac[a[i]])
                 l[i] = max(l[i], vis[it]), vis[it] = i;
         memset(vis, 0x3f, sizeof vis);
         for (int i = n; i >= ; --i)
             for (auto it : fac[a[i]])
                 r[i] = min(r[i], vis[it]), vis[it] = i;
         //for (int i = 1; i <= n; ++i) printf("%d %d\n", l[i], r[i]);
         //for (int i = 1; i <= n; ++i) for (int j = 0, len = fac[a[i]].size(); j < len; ++j) printf("%d%c", fac[a[i]][j], " \n"[j == len - 1]);
         for (int i = ; i <= n; ++i)
             v[l[i]].emplace_back(r[i], i);
         for (int qq = , l, r; qq <= q; ++qq)
         {
             scanf("%d%d", &l, &r);
             qv[l].emplace_back(r, qq);
         }
         SEG::build(, , n);
         for (int i = ; i <= n; ++i)
         {
             for (auto it : v[i - ])
                 SEG::update(, , n, it.second, it.first - , );
             for (auto it : qv[i])
                 ans[it.second] = SEG::query(, , n, it.first);
             SEG::update(, , n, i, r[i] - , -);
         }
         for (int i = ; i <= q; ++i) printf("%d\n", ans[i]);
     }
     return ;
 }

I：

Solved.

特判$n == 0 并且 m == 0 的情况$

其他情况下，因为$k >= |n - m| 先手的人只需要去掉石子使得两堆相同，然后跟着另一个人取就可以必胜$

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 ll n, m, k;

 int main()
 {
     while(~scanf("%lld %lld %lld", &n, &m, &k))
     {
         if(n ==  || m == ) puts("LAOZI CHUI SI NI!");
         else puts("HAI YOU SEI!");
     }
     return ;
 }

J：

Solved.

考虑 末尾两位为00, 25, 50, 75四种状态

 #include<bits/stdc++.h>

 using namespace std;

 const int INF = 0x3f3f3f3f;
 const int maxn = 1e5 + ;

 char str[maxn];

 int main()
 {
     while(~scanf("%s", str + ))
     {
         int ans = INF;
         int len = strlen(str + );

         //
         int tmp1 = -, tmp2 = -;
         for(int i = len; i >= ; --i)
         {
             if(str[i] == '')
             {
                 if(tmp1 == -) tmp1 = len - i;
                 else if(tmp2 == -)
                 {
                     tmp2 = len - i - ;
                     break;
                 }
             }
         }
         if(tmp1 != - && tmp2 != -) ans = min(ans, tmp1 + tmp2);

         //
         tmp1 = -, tmp2 = -;
         for(int i = len; i >= ; --i)
         {
             if(str[i] == '')
             {
                 if(tmp1 == -) tmp1 = len - i;
             }
             if(str[i] == '')
             {
                 if(tmp2 == -) tmp2 = len - i;
             }
         }

         if(tmp1 != - && tmp2 != -)
         {
             if(tmp1 < tmp2)
             {
                 ans = min(ans, tmp1 + tmp2 - );
             }
             else
             {
                 ans = min(ans, tmp1 + tmp2);
             }
         }

         //
         tmp1 = -, tmp2 = -;
         for(int i = len; i >= ; --i)
         {
             if(str[i] == '')
             {
                 if(tmp1 == -) tmp1 = len - i;
             }
             if(str[i] == '')
             {
                 if(tmp2 == -) tmp2 = len - i;
             }
         }

         if(tmp1 != - && tmp2 != -)
         {
             if(tmp1 < tmp2)
             {
                 ans = min(ans, tmp1 + tmp2 - );
             }
             else
             {
                 ans = min(ans, tmp1 + tmp2);
             }
         }
         //
         tmp1 = -, tmp2 = -;
         for(int i = len; i >= ; --i)
         {
             if(str[i] == '')
             {
                 if(tmp1 == -) tmp1 = len - i;
             }
             if(str[i] == '')
             {
                 if(tmp2 == -) tmp2 = len - i;
             }
         }
         if(tmp1 != - && tmp2 != -)
         {
             if(tmp1 < tmp2)
             {
                 ans = min(ans, tmp1 + tmp2 - );
             }
             else
             {
                 ans = min(ans, tmp1 + tmp2);
             }
         }
         if(ans == INF) ans = -;
         printf("%d\n", ans);
     }
     return ;
 }