2015 Multi-University Training Contest 3

好吧，脑残的孩子仅仅能慢慢来。

。

---

1002、RGCDQ

题目传送：RGCDQ

人脑残到写了个线段树。。。然后跪啦好久。

。。然后就滚去睡觉了

事实上这个题不须要用到线段树，仅仅须要维护前缀和就ok了，

由于f的值非常小。所以能够暴力筛出来。然后用一个s[i][j]数组存储，s[i][j]代表前i个f值中有多少个等于j的

线段树仅仅有要更新的时候才拿出来！

不更新的时候要先想到前缀和。

AC代码（前缀和）：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 1000005;

int f[maxn];
int s[maxn][8];

void init() {
    for(int i = 2; i < maxn; i ++) {
        if(f[i] == 0)
        for(int j = i; j < maxn; j += i) {
            f[j] ++;
        }
    }
    for(int i = 2; i < maxn; i ++) {
        for(int j = 1; j < 8; j ++) {
            s[i][j] += s[i-1][j];
        }
        s[i][f[i]] ++;
    }
}

int main() {
    init();
    int T;
    scanf("%d", &T);
    while(T --) {
        int L, R;
        scanf("%d %d", &L, &R);
        int t1 = s[R][1] - s[L-1][1];
        int t2 = s[R][2] - s[L-1][2];
        int t3 = s[R][3] - s[L-1][3];
        int t4 = s[R][4] - s[L-1][4];
        int t5 = s[R][5] - s[L-1][5];
        int t6 = s[R][6] - s[L-1][6];
        int t7 = s[R][7] - s[L-1][7];
        if(t7 >= 2) {
            printf("7
");
            continue;
        }
        if(t6 >= 2) {
            printf("6
");
            continue;
        }
        if(t5 >= 2) {
            printf("5
");
            continue;
        }
        if(t4 >= 2) {
            printf("4
");
            continue;
        }
        if(t3 >= 2 || (t3 >= 1 && t6 >= 1)) {
            printf("3
");
            continue;
        }
        if(t2 >= 2 || (t2 >= 1 && t4 >= 1) || (t2 >= 1 && t6 >= 1) || (t4 >= 1 && t6 >= 1)) {
            printf("2
");
            continue;
        }
        printf("1
");
    }
    return 0;
}

AC代码（线段树）：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <ctime>
#define LL long long
#define INF 0x7fffffff
#define max(a, b) a > b ? a : b
using namespace std;

const int maxn = 1000005;

int biao[maxn];

void init() {
    for(int i = 2; i < maxn; i ++) {
        if(biao[i] == 0)
        for(int j = i; j < maxn; j += i) {
            biao[j] ++;
        }
    }
}

int x2[maxn << 2];
int x3[maxn << 2];
int x4[maxn << 2];
int x5[maxn << 2];
int x6[maxn << 2];
int x7[maxn << 2];

void build(int l, int r, int rt) {  
    if(l == r) {  
        switch(biao[l]) {
            case 2: x2[rt] = 1; break;
            case 3:    x3[rt] = 1; break;
            case 4:    x4[rt] = 1; break;
            case 5:    x5[rt] = 1; break;
            case 6:    x6[rt] = 1; break;
            case 7:    x7[rt] = 1; break;
        }
        return;  
    }  
    int mid = (l + r) >> 1;  
    build(l, mid, rt << 1);  
    build(mid + 1, r, rt << 1 | 1);  
    x2[rt] = x2[rt << 1] + x2[rt << 1 | 1];
    x3[rt] = x3[rt << 1] + x3[rt << 1 | 1];
    x4[rt] = x4[rt << 1] + x4[rt << 1 | 1];
    x5[rt] = x5[rt << 1] + x5[rt << 1 | 1];
    x6[rt] = x6[rt << 1] + x6[rt << 1 | 1];
    x7[rt] = x7[rt << 1] + x7[rt << 1 | 1];
}  

int query2(int L, int R, int l, int r, int rt) {  
    if(L <= l && r <= R) {  
        return x2[rt];  
    }
    int ret = 0;
    int mid = (l + r) >> 1;  
    if(L <= mid) ret += query2(L, R, l, mid, rt << 1);
    if(R >= mid + 1) ret += query2(L, R, mid + 1, r, rt << 1 | 1);
    return ret;
}  

int query3(int L, int R, int l, int r, int rt) {  
    if(L <= l && r <= R) {  
        return x3[rt];  
    }
    int ret = 0;
    int mid = (l + r) >> 1;  
    if(L <= mid) ret += query3(L, R, l, mid, rt << 1);
    if(R >= mid + 1) ret += query3(L, R, mid + 1, r, rt << 1 | 1);
    return ret;
} 

int query4(int L, int R, int l, int r, int rt) {  
    if(L <= l && r <= R) {  
        return x4[rt];  
    }
    int ret = 0;
    int mid = (l + r) >> 1;  
    if(L <= mid) ret += query4(L, R, l, mid, rt << 1);
    if(R >= mid + 1) ret += query4(L, R, mid + 1, r, rt << 1 | 1);
    return ret;
}  

int query5(int L, int R, int l, int r, int rt) {  
    if(L <= l && r <= R) {  
        return x5[rt];  
    }
    int ret = 0;
    int mid = (l + r) >> 1;  
    if(L <= mid) ret += query5(L, R, l, mid, rt << 1);
    if(R >= mid + 1) ret += query5(L, R, mid + 1, r, rt << 1 | 1);
    return ret;
}  

int query6(int L, int R, int l, int r, int rt) {  
    if(L <= l && r <= R) {  
        return x6[rt];  
    }
    int ret = 0;
    int mid = (l + r) >> 1;  
    if(L <= mid) ret += query6(L, R, l, mid, rt << 1);
    if(R >= mid + 1) ret += query6(L, R, mid + 1, r, rt << 1 | 1);
    return ret;
}

int query7(int L, int R, int l, int r, int rt) {  
    if(L <= l && r <= R) {  
        return x7[rt];  
    }
    int ret = 0;
    int mid = (l + r) >> 1;  
    if(L <= mid) ret += query7(L, R, l, mid, rt << 1);
    if(R >= mid + 1) ret += query7(L, R, mid + 1, r, rt << 1 | 1);
    return ret;
}

int main() {
    init();
    build(1, maxn, 1);
    int T;
    scanf("%d", &T);
    while(T --) {
        int L, R;
        scanf("%d %d", &L, &R);
        int t7 = query7(L, R, 1, maxn, 1);
        if(t7 >= 2) {
            printf("7
");
            continue;
        }
        int t6 = query6(L, R, 1, maxn, 1);
        if(t6 >= 2) {
            printf("6
");
            continue;
        }
        int t5 = query5(L, R, 1, maxn, 1);
        if(t5 >= 2) {
            printf("5
");
            continue;
        }
        int t4 = query4(L, R, 1, maxn, 1);
        if(t4 >= 2) {
            printf("4
");
            continue;
        }
        int t3 = query3(L, R, 1, maxn, 1);
        if(t3 >= 2 || (t3 >= 1 && t6 >= 1)) {
            printf("3
");
            continue;
        }
        int t2 = query2(L, R, 1, maxn, 1);
        if(t2 >= 2 || (t2 >= 1 && t4 >= 1) || (t2 >= 1 && t6 >= 1) || (t4 >= 1 && t6 >= 1)) {
            printf("2
");
            continue;
        }

        printf("1
");
    }
    return 0;
}

---

1011、Work

题目传送：Work

简单题。仅仅需建立起有向图。然后遍历就可以，遍历的时候累加一下

AC代码：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 105;
vector<int> mp[105];

int ans[105];

void dfs(int x) {
    int m = mp[x].size();
    for(int i = 0; i < m; i ++) {
        ans[mp[x][i]] ++;
        dfs(mp[x][i]);
    }
}

int main() {
    int n, k;
    while(scanf("%d %d", &n, &k) != EOF) {
        for(int i = 1; i < n; i ++) {
            int u, v;
            scanf("%d %d", &u, &v);
            mp[v].push_back(u);
        }

        memset(ans, 0, sizeof(ans));

        for(int i = 1; i <= n; i ++) {
            dfs(i);
        }

        int cnt = 0;
        for(int i = 1; i <= n; i ++) {
            if(ans[i] == k) cnt ++;
        }

        printf("%d
", cnt);

        for(int i = 0; i <= n; i ++) {
            mp[i].clear();
        }
    }
    return 0;
}