2015 Multi-University Training Contest 10（hdu 5406 - hdu 5416）

1.CRB and Apple

2.CRB and Candies

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

typedef long long ll;
const int MOD = 1000000007;
const int N = 1000005;
ll f
;
int nprime;
int vis
;
int prime[80000];

void getprime(){
    nprime = 0;
    memset(vis,0,sizeof(vis));
    memset(prime,0,sizeof(prime));
    for(int i = 2; i <= N-5; i++){
        int t = (N-5)/i;
        for(int j = 2; j <= t; j++){
            vis[i*j] = 1;
        }
    }
    for(int i = 2; i <= N-5; i++){
        if(!vis[i])
            prime[nprime++] = i;
    }
    memset(vis,0,sizeof(vis));
    for(int i = 0; i < nprime; i++){
        ll a = prime[i];
        ll b = a;
        for(; a < N; a*=b)
            vis[a] = b;
    }
}

void init(){
    getprime();
    f[1] = 1;
    for(int i = 2; i <= N-4; i++){
        if(vis[i])
            f[i] = f[i-1]*vis[i]%MOD;
        else
            f[i] = f[i-1];
        f[i] %= MOD;
    }
}
/*
ll extend_gcd(ll a,ll b,ll &x,ll &y){
    if(a == 0 && b == 0)
        return -1;//无最大公约数
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    ll d = extend_gcd(b,a%b,y,x);
    y -= a/b*x;
    return d;
}

//*********求逆元素*******************
//ax = 1(mod n)
ll mod_reverse(ll a,ll n)
{
    ll x,y;
    ll d = extend_gcd(a,n,x,y);
    if(d == 1)
        return (x%n+n)%n;
    else
        return -1;
}
*/

ll pow_mod(ll x, int n) {
    ll ret = 1;
    while (n) {
        if (n&1)
            ret = ret * x % MOD;
        x = x * x % MOD;
        n >>= 1;
    }
    return ret;
}

ll mod_reverse(ll x) {
    return pow_mod(x, MOD-2);
}

int main(){
    init();
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        //ll ni = mod_reverse(n+1,MOD);
        ll ni = mod_reverse(n+1);
        printf("%lld\n",f[n+1]*ni%MOD);
    }
    return 0;
}

3.CRB and Farm

4.CRB and Graph

5.CRB and His Birthday

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int dp[2005];
int val[1005];
int a[1005],b[1005];

int main(){
    int T;
    int n,m;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&m,&n);
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= n; i++)
            scanf("%d%d%d",&val[i],&a[i],&b[i]);
        for(int i = 1; i <= n; i++)
            for(int j = m; j >= val[i]; j--)
                dp[j] = max(dp[j],dp[j-val[i]]+a[i]+b[i]);
        for(int i = 1; i <= n; i++)
            for(int j = val[i]; j <= m; j++)
                dp[j] = max(dp[j],dp[j-val[i]]+a[i]);
        printf("%d\n",dp[m]);
    }
    return 0;
}

6.CRB and Puzzle

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int MOD = 2015;
struct matrix//矩阵
{
    int m[55][55];
    matrix(){
        memset(m,0,sizeof(m));
    }
};
int n,m;

void debug(matrix a){
    for(int i = 1; i <= n+1;i++){
        for(int j = 1; j <= n+1; j++){
            cout<<a.m[i][j];
        }
        cout<<endl;
    }
}

matrix multi(matrix a, matrix b)
{
    matrix tmp;
    for(int i = 0; i < 55; ++i)
    {
        for(int j = 0; j < 55; ++j)
        {
            for(int k = 0; k < 55; ++k)
                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
        }
    }
    return tmp;
}

int fast_mod(matrix ans,matrix base, int m)  // 求矩阵 base 的  n 次幂
{
    while(m)
    {
        if(m & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t
            ans = multi(ans, base);
        base = multi(base, base);
        m >>= 1;
    }
    return ans.m[1][n+1];
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        matrix ans,base;
        for(int i = 1; i <= n+1; i++)
           base.m[i][n + 1] = 1;
        int x,xx;
        for(int i = 1; i <= n; i++){
            scanf("%d",&x);
            for(int j = 1; j <= x; j++){
                scanf("%d",&xx);
                base.m[i][xx] = 1;
            }
        }
        for(int i = 1; i <= n+1; i++)
            ans.m[1][i] = 1;
        if(m == 1)
            printf("%d\n",n+1);
        else
            printf("%d\n",fast_mod(ans,base,m));
    }
    return 0;
}

7.CRB and Queries

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
#define maxn 300007
int tree[maxn];
void add(int p,int n){
    for(;p<maxn;p+=p&(-p))
        tree[p]+=n;
}
int query(int p){
    int ans = 0;
    for(;p>0;p-=p&(-p))
        ans += tree[p];
    return ans;
}

struct Node{
    int l,r,k,ty,ans;
};
Node p[maxn];
int id1[maxn],id2[maxn];
void CDQ(int L,int R,int low,int high){
    if(R < L) return ;
    if(low == high ){
        for(;L<=R;L++){
            p[id1[L]].ans = low;
        }
        return ;
    }
    int mid = (low+high)/2,l=L,r=R,k,u;
    for(int i = L;i <= R; i++){
        u = id1[i];
       if(p[u].ty == 2){
            k = query(p[u].r) - query(p[u].l-1);
            if(k >= p[u].k) id2[l++] = u;
            else {
                p[u].k -= k;
                id2[r--] = u;
            }
        }
        else if(p[u].k <= mid){
            add(p[u].l,p[u].ty);
            id2[l++] = u;
        }
        else id2[r--] =  u;
    }

    for(int i = L; i <= R; i++){
        u = id1[i];
        if(p[u].ty != 2 && p[u].k <= mid) add(p[u].l,-p[u].ty);
    }
    for(k=L;k<l;k++)
        id1[k] = id2[k];
    for(r=R;k<=R;k++)
        id1[k] = id2[r--];
    CDQ(L,l-1,low,mid);
    CDQ(l,R,mid+1,high);
}

int num[maxn];

int main(){
    int n,q,t,cnt;
    memset(tree,0,sizeof(tree));
    while(scanf("%d",&n)!=EOF){
        for(cnt=0;cnt<n;cnt++){
            scanf("%d",&p[cnt].k);
            p[cnt].ty = 1;
            p[cnt].l = cnt+1;
            num[cnt+1] = p[cnt].k;
        }
        scanf("%d",&q);
        int ty,l,v;
        for(int i = 0;i < q; i++,cnt++){
            scanf("%d",&p[cnt].ty);
            if(p[cnt].ty == 1){
                scanf("%d%d",&l,&v);
                p[cnt].ty = -1;
                p[cnt].k = num[l];
                p[cnt].l = l;
                cnt++;
                num[l] = v;
                p[cnt].ty = 1;
                p[cnt].k = v;
                p[cnt].l = l;
            }
            else {
                scanf("%d%d%d",&p[cnt].l,&p[cnt].r,&p[cnt].k);
            }
        }
        for(int i = 0;i < cnt; i++)
            id1[i] = i;
        CDQ(0,cnt-1,0,1000000000);
        for(int i = 0;i < cnt; i++){
            if(p[i].ty == 2) printf("%d\n",p[i].ans);
        }
    }
    return 0;
}

8.CRB and Roads

9.CRB and String

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 100005;
char s
,t
;

int solve(){
    int i,j;
    int l1 = strlen(s);
    int l2 = strlen(t);
    for(i = 1; i < l2; i++)
        if(t[i] != t[0])
            break;//找到t串的第一个不连续的位置
    for(j = 0; j < i; j++)//看s的前i个子串是否连续
        if(s[j] != t[0])
            return 0;
    for(; j<l1;){
        for(; i<l2; i++){//找到下一个和s相等的地方
            if(t[i] == s[j])
                break;
        }
        if(i == l2)
            return 0;//如果t找完了还没跳出证明s不是t的子串
        i++;
        j++;
    }
    return 1;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s%s",s,t);
        if(solve())
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

10.CRB and Substrings

11.CRB and Tree

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

typedef long long ll;
const int N = 100005;
const int maxn = N<<1;
struct node{
    int x,w;
};
vector<node> edge
;
int vis
;
ll cnt[maxn];

void dfs(int u,int val){
    vis[u] = 1;
    if(u^1)
        cnt[val]++;
    int l = edge[u].size();
    for(int i = 0; i < l; i++){
        node tmp = edge[u][i];
        if(!vis[tmp.x]){
            dfs(tmp.x,tmp.w^val);
        }
    }
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,q;
        scanf("%d",&n);
        for(int i = 1; i <= n; i++)
            edge[i].clear();
        int u,v,w;
        for(int i = 1; i < n; i++){
            scanf("%d%d%d",&u,&v,&w);
            edge[u].push_back(node{v,w});
            edge[v].push_back(node{u,w});
        }
        memset(vis,0,sizeof(vis));
        memset(cnt,0,sizeof(cnt));
        dfs(1,0);

        scanf("%d",&q);
        while(q--){
            scanf("%d",&w);
            ll ans = 0;
            if(w == 0){
                ans += n+cnt[0];
                for(int i = 0; i < maxn; i++)
                    ans += cnt[i] * (cnt[i] - 1) / 2;
            }
            else{
                for(int i = 0; i < maxn; i++)
                    ans += cnt[i] * (cnt[i ^ w]);
                ans >>= 1;
                ans += cnt[w];
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}