Codeforces Round #232 (Div. 2)

Problems

 

 

#Name  A

On Segment's Own Points

standard input/output

1 s, 256 MB

  x1657B

On Corruption and Numbers

standard input/output

1 s, 256 MB

  x925C

On Number of Decompositions into Multipliers

standard input/output

1 s, 256 MB

  x181D

On Sum of Fractions

standard input/output

2 s, 256 MB

  x134E

On Changing Tree

standard input/output

2 s, 256 MB

  x34

A题：n个区间，你可以选择第一个区间上的位置，后面n-1行是被占掉的区间，求你最多能占多长的区间。

思路：n才100，直接暴力，把出现过的区间标记掉，最后去遍历一遍即可。

B题：你有l-r的硬币，要组合出x的钱，问能否组合。

思路：可以的区间为1*[l,r], 2 * [l,r], 3 * [l,r]....直到后面区间重合了之后都是一直可以的，所以用x / l求出i。然后乘上r判断n在不在区间内即可。

C题：m是a1*a2*a3..*an。问m有几种分解成n个数相乘的不同方法。

思路：先分解所有a的分解成质因子，然后等同于把质因子放入n个位置去，用隔板法，每个质因子的方法为C(n - 1 + k) (n - 1)种，k为该质因子个数。

D题：求出题目给定公式值。

思路：先推公式1/u(i) * 1/v(i) = 1/(v(i) - u(i)) * (1/v(i) - 1/u(i))。如此一来前面每一项等于(1/2 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7).....(1/m - 1/n） = 1/2 - 1/n。然后关键就变成找出n的上下质数，这步用暴力枚举，直到是质数为止。然后求出总和即可。

E题：n个点的有根树，根为1，操作1在v结点添加，距离为i的子节点添加值为x - i * k。2为询问。

思路：树状数组，在添加的时候，先假设是从根添加，这样要多添加k * dep[v]。然后开2个树状数组一个记录sum和一个记录k。这样一来最后答案变为

sum - k * dep[v];

代码：

A：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 105;int n, i, vis[N], l, r, ll, rr;int main() {    scanf("%d", &n);    scanf("%d%d", &ll, &rr);    for (i = 2; i <= n; i++) {        scanf("%d%d", &l, &r);        for (int j = l; j < r; j++)            vis[j] = 1;    }    int ans = 0;    for (i = ll; i < rr; i++) {        if (!vis[i])            ans++;    }    printf("%d\n", ans);    return 0;}

B：

#include <stdio.h>#include <string.h>int t;__int64 n, l, r, i;bool solve() {    if (n < l) return false;    __int64 k = n / l;    if (n <= r * k) return true;    return false;}int main() {    scanf("%d", &t);    while (t--) {        scanf("%I64d%I64d%I64d", &n, &l, &r);        printf("%s\n", solve()?"Yes":"No");    }    return 0;}

C：

#include <stdio.h>#include <string.h>#include <math.h>#include <map>using namespace std;const int MOD = 1000000007;const int N = 505;const int MAXN = 20005;int n, a, cnt = 0, num[MAXN], c[20005][1005];map<int ,int> v;void getnum(int x) {for (int i = 2; i * i <= x; i++) {while (x % i == 0) {if (v.count(i)) {num[v[i]]++;x /= i;}else {v[i] = ++cnt;num[v[i]]++;x /= i;}}}if (x == 1) return;if (v.count(x)) {num[v[x]]++;}else {v[x] = ++cnt;num[v[x]]++;}}void init() {c[0][0] = 1;for (int i = 1; i <= 17000; i++)for (int j = 0; j <= i && j <= 1000; j++)if (j == 0 || j == i) c[i][j] = 1;else c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;}__int64 cal(int k) {return c[k + n - 1][n - 1];}__int64 solve() {__int64 ans = 1;for (int i = 1; i <= cnt; i++) ans = ans * cal(num[i]) % MOD;return ans;}int main() {init();scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &a);getnum(a);}printf("%I64d\n", solve());return 0;}

D：

#include <stdio.h>#include <string.h>const int MAXN = 100005;int t;__int64 n, l, r, prime[MAXN], vis[MAXN], pn = 0;void init() {for (int i = 2; i <= 100000; i++) {if (vis[i]) continue;prime[pn++] = i;for (int j = i; j <= 100000; j += i)vis[j] = 1;}}bool judge(int x) {for (int i = 0; i < pn; i++) {if (x % prime[i] == 0 && x != prime[i])return false;}return true;}__int64 find_l(__int64 x) {while (1) {if (judge(x))return x;x--;}}__int64 find_r(__int64 x) {x++;while (1) {if (judge(x))return x;x++;}}__int64 gcd(__int64 a, __int64 b) {if (b == 0) return a;return gcd(b, a%b);}void print(__int64 l, __int64 r) {__int64 zi = (l - 2) * r + (n - l + 1) * 2, mu = l * r * 2;printf("%I64d/%I64d\n", zi / gcd(zi, mu), mu / gcd(zi, mu));}int main() {init();scanf("%d", &t);while (t--) {scanf("%d", &n);l = find_l(n); r = find_r(n);print(l, r);}return 0;}

E：

#include <stdio.h>#include <string.h>#include <vector>using namespace std;const int N = 300005;const int MOD = 1000000007;int n, Q, i, nod, vis[N];__int64 kbit[N], sbit[N], cnt = 0, l[N], r[N], dep[N];vector<int> g[N];void dfs(int u, __int64 d) {    vis[u] = 1; dep[u] = d;    cnt++; l[u] = cnt;    for (int i = 0; i < g[u].size(); i++) {        int v = g[u][i];        if (vis[v]) continue;        dfs(v, d + 1);    }    r[u] = cnt;}void Is(__int64 value, int x, __int64 *num) {    while (x <= N) {        num[x] = (num[x] + value) % MOD;        x += (x&(-x));    }}__int64 Sum(int x, __int64 *num) {    __int64 ans = 0;    while (x > 0) {        ans = (ans + num[x]) % MOD;        x -= (x&(-x));    }    return ans;}int main() {    scanf("%d", &n);    for (i = 2; i <= n; i++) {        scanf("%d", &nod);        g[nod].push_back(i);    }    dfs(1, 0);    scanf("%d", &Q);    while (Q--) {        int type, v;        scanf("%d%d", &type, &v);        if (type == 1) {            __int64 x, k;            scanf("%I64d%I64d", &x, &k);            __int64 val = (x + dep[v] * k) % MOD;            Is(k, l[v], kbit);            Is(-k, r[v] + 1, kbit);            Is(val, l[v], sbit);            Is(-val, r[v] + 1, sbit);        }        else printf("%I64d\n", ((Sum(l[v], sbit) - dep[v] * Sum(l[v], kbit)) % MOD + MOD) % MOD);    }    return 0;}