Codeforces Round #478 (Div. 2)

题目链接：http://codeforces.com/contest/975

 

A. Aramic script

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

In Aramic language words can only represent objects.

Words in Aramic have special properties:

• A word is a root if it does not contain the same letter more than once.
• A root and all its permutations represent the same object.
• The root $\mathrm{xx\; of\; a\; word\; yy\; is\; the\; word\; that\; contains\; all\; letters\; that\; appear\; in\; yy\; in\; a\; way\; that\; each\; letter\; appears\; once.\; For\; example,\; the\; root\; of\; "aaaa",\; "aa",\; "aaa"\; is\; "a",\; the\; root\; of\; "aabb",\; "bab",\; "baabb",\; "ab"\; is\; "ab".}$
• Any word in Aramic represents the same object as its root.

You have an ancient script in Aramic. What is the number of different objects mentioned in the script?

Input

The first line contains one integer $\mathrm{nn\; (1\le n\le 1031\le n\le 103) —\; the\; number\; of\; words\; in\; the\; script.}$

The second line contains $\mathrm{nn\; words\; s1,s2,\dots ,sns1,s2,\dots ,sn —\; the\; script\; itself.\; The\; length\; of\; each\; string\; does\; not\; exceed\; 103103.}$

It is guaranteed that all characters of the strings are small latin letters.

Output

Output one integer — the number of different objects mentioned in the given ancient Aramic script.

Examples

Input

Copy

5
a aa aaa ab abb

Output

Copy

2

Input

Copy

3
amer arem mrea

Output

Copy

1

Note

In the first test, there are two objects mentioned. The roots that represent them are "a","ab".

In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".

题意：给你n个字符串，然后把每个字符串中出现的字母取出来（构成的新字符串忽略顺序记为根），问这n个字符串中不同根的个数。

思路：用一个数组a记录每个字符串中出现的字母，然后再把它按照字典序构成一个新的字符串用set来维护，结果就是set所含的元素数。

代码实现如下：

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

int n;
int a[26];
string s[1007], ss[1007];

int main() {
    cin >>n;
    for(int i = 0; i < n;i++) {
        getchar();
        cin >>s[i];
        ss[i].clear();
        memset(a, 0, sizeof(a));
        int len = s[i].size();
        for(int j = 0; j < len; j++) {
            int t = s[i][j] - 'a';
            a[t]++;
        }
        for(int j = 0; j < 26; j++) {
            if(a[j]) ss[i] += (j + 'a');
        }
    }
    set<string> m;
    for(int i = 0; i < n; i++) {
        if(!m.count(ss[i])) {
            m.insert(ss[i]);
        }
    }
    cout <<m.size() <<endl;
    return 0;
}

 

 

B. Mancala

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.

Initially, each hole has ${\mathrm{aiai\; stones.\; When\; a\; player\; makes\; a\; move,\; he\; chooses\; a\; hole\; which\; contains\; a\; positive\; number\; of\; stones.\; He\; takes\; all\; the\; stones\; inside\; it\; and\; then\; redistributes\; these\; stones\; one\; by\; one\; in\; the\; next\; holes\; in\; a\; counter-clockwise\; direction.}}^{}$

Note that the counter-clockwise order means if the player takes the stones from hole $\mathrm{ii,\; he\; will\; put\; one\; stone\; in\; the\; (i+1)(i+1)-th\; hole,\; then\; in\; the\; (i+2)(i+2)-th,\; etc.\; If\; he\; puts\; a\; stone\; in\; the\; 1414-th\; hole,\; the\; next\; one\; will\; be\; put\; in\; the\; first\; hole.}$

After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.

Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.

Input

The only line contains 14 integers ${\mathrm{a1,a2,\dots ,a14a1,a2,\dots ,a14\; (0\le ai\le 1090\le ai\le 109) —\; the\; number\; of\; stones\; in\; each\; hole.}}^{}$

It is guaranteed that for any $\mathrm{ii\; (1\le i\le 141\le i\le 14)\; aiai\; is\; either\; zero\; or\; odd,\; and\; there\; is\; at\; least\; one\; stone\; in\; the\; board.}$

Output

Output one integer, the maximum possible score after one move.

Examples

Input

Copy

0 1 1 0 0 0 0 0 0 7 0 0 0 0

Output

Copy

4

Input

Copy

5 1 1 1 1 0 0 0 0 0 0 0 0 0

Output

Copy

8

Note

In the first test case the board after the move from the hole with $\mathrm{77\; stones\; will\; look\; like\; 1\; 2\; 2\; 0\; 0\; 0\; 0\; 0\; 0\; 0\; 1\; 1\; 1\; 1.\; Then\; the\; player\; collects\; the\; even\; numbers\; and\; ends\; up\; with\; a\; score\; equal\; to\; 44.}$

 

 代码实现如下：

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

long long a[20], b[20];

int main() {
    for(int i = 1; i <= 14; i++) {
        cin >> a[i];
    }
    long long ans = 0, sum;
    for(int i = 1; i <= 14; i++) {
        sum = 0;
        memset(b, 0, sizeof(b));
        if(a[i] <= 0)
            continue;
        for(int j = 1; j <= 14; j++) {
            if(i != j) {
                b[j] = a[j];
            }
        }
        int n = a[i] / 14, m = a[i] % 14;
        for(int j = 1; j <= 14; j++) b[j] += n;
        for(int j = i + 1; j <= 14 && m > 0; j++, m--) {
            b[j] ++;
        }
        for(int j = 1; m > 0; j++, m--) {
            b[j] ++;
        }
        for(int j = 1; j <= 14; j++) {
            if(b[j] % 2 == 0) {
                sum += b[j];
            }
        }
        ans = max(ans, sum);
    }
    cout << ans << endl;
    return 0;
}

 

 

C. Valhalla Siege

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Ivar the Boneless is a great leader. He is trying to capture Kattegat from Lagertha. The war has begun and wave after wave Ivar's warriors are falling in battle.

Ivar has $\mathrm{nn\; warriors,\; he\; places\; them\; on\; a\; straight\; line\; in\; front\; of\; the\; main\; gate,\; in\; a\; way\; that\; the\; ii\; -th\; warrior\; stands\; right\; after\; (i-1)(i-1)\; -th\; warrior.\; The\; first\; warrior\; leads\; the\; attack.}$

Each attacker can take up to ${\mathrm{aiai\; arrows\; before\; he\; falls\; to\; the\; ground,\; where\; aiai\; is\; the\; ii\; -th\; warrior\text{'}s\; strength.}}^{}$

Lagertha orders her warriors to shoot ${\mathrm{kiki\; arrows\; during\; the\; ii\; -th\; minute,\; the\; arrows\; one\; by\; one\; hit\; the\; first\; still\; standing\; warrior.\; After\; all\; Ivar\text{'}s\; warriors\; fall\; and\; all\; the\; currently\; flying\; arrows\; fly\; by,\; Thor\; smashes\; his\; hammer\; and\; all\; Ivar\text{'}s\; warriors\; get\; their\; previous\; strengths\; back\; and\; stand\; up\; to\; fight\; again.\; In\; other\; words,\; if\; all\; warriors\; die\; in\; minute\; tt\; ,\; they\; will\; all\; be\; standing\; to\; fight\; at\; the\; end\; of\; minute\; tt\; .}}^{}$

The battle will last for $\mathrm{qq\; minutes,\; after\; each\; minute\; you\; should\; tell\; Ivar\; what\; is\; the\; number\; of\; his\; standing\; warriors.}$

Input

The first line contains two integers $\mathrm{nn\; and\; qq\; (1\le n,q\le 2000001\le n,q\le 200000\; ) —\; the\; number\; of\; warriors\; and\; the\; number\; of\; minutes\; in\; the\; battle.}$

The second line contains $\mathrm{nn\; integers\; a1,a2,\dots ,ana1,a2,\dots ,an\; (1\le ai\le 1091\le ai\le 109\; )\; that\; represent\; the\; warriors\text{'}\; strengths.}$

The third line contains $\mathrm{qq\; integers\; k1,k2,\dots ,kqk1,k2,\dots ,kq\; (1\le ki\le 10141\le ki\le 1014\; ),\; the\; ii\; -th\; of\; them\; represents\; Lagertha\text{'}s\; order\; at\; the\; ii\; -th\; minute:\; kiki\; arrows\; will\; attack\; the\; warriors.}$

Output

Output $\mathrm{qq\; lines,\; the\; ii\; -th\; of\; them\; is\; the\; number\; of\; standing\; warriors\; after\; the\; ii\; -th\; minute.}$

Examples

Input

Copy

5 5
1 2 1 2 1
3 10 1 1 1

Output

Copy

3
5
4
4
3

Input

Copy

4 4
1 2 3 4
9 1 10 6

Output

Copy

1
4
4
1

Note

In the first example:

• after the 1-st minute, the 1-st and 2-nd warriors die.
• after the 2-nd minute all warriors die (and all arrows left over are wasted), then they will be revived thus answer is 5 — all warriors are alive.
• after the 3-rd minute, the 1-st warrior dies.
• after the 4-th minute, the 2-nd warrior takes a hit and his strength decreases by 1.
• after the 5-th minute, the 2-nd warrior dies.

题意：人肉盾请了解一下，第i个人可以抵挡ai只弓箭（看作hp），第i分钟地方射出ki只箭， 战争持续了q分钟。问每分钟还有多少人还活着（注意，当所有人都死亡时，将会集体复活，雾）。

思路：用前缀和记录前i个人的总hp，然后用t来记录前i分钟的弓箭数，当t>=sum[n]时将t设为0，然后进行二分即可。

代码实现如下：

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

const int maxn = 2e5 + 7;
int n, q;
long long x;
long long a[maxn], sum[maxn];

int main() {
    cin >>n >>q;
    sum[0] = 0;
    for(int i= 1; i<=n;i++) {
        cin >>a[i];
        sum[i] = sum[i-1] + a[i];
    }
    long long  t = 0;
    while(q--) {
        cin >>x;
        t += x;
        if(t >= sum[n]) t = 0;
        int pos = upper_bound(sum,sum + n + 1, t) - sum - 1;
        //cout <<pos <<endl <<endl;
        cout <<n - pos <<endl;
    }
    return 0;
}

D. Ghosts

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Ghosts live in harmony and peace, they travel the space without any purpose other than scare whoever stands in their way.

There are $\mathrm{nn\; ghosts\; in\; the\; universe,\; they\; move\; in\; the\; OXYOXY\; plane,\; each\; one\; of\; them\; has\; its\; own\; velocity\; that\; does\; not\; change\; in\; time:\; \to V=Vx\to i+Vy\to jV\to =Vxi\to +Vyj\to \; where\; VxVx\; is\; its\; speed\; on\; the\; xx-axis\; and\; VyVy\; is\; on\; the\; yy-axis.}$

A ghost $\mathrm{ii\; has\; experience\; value\; EXiEXi,\; which\; represent\; how\; many\; ghosts\; tried\; to\; scare\; him\; in\; his\; past.\; Two\; ghosts\; scare\; each\; other\; if\; they\; were\; in\; the\; same\; cartesian\; point\; at\; a\; moment\; of\; time.}$

As the ghosts move with constant speed, after some moment of time there will be no further scaring (what a relief!) and the experience of ghost kind $\mathrm{GX=\sum ni=1EXiGX=\sum i=1nEXi\; will\; never\; increase.}$

Tameem is a red giant, he took a picture of the cartesian plane at a certain moment of time $\mathrm{TT,\; and\; magically\; all\; the\; ghosts\; were\; aligned\; on\; a\; line\; of\; the\; form\; y=a\cdot x+by=a\cdot x+b.\; You\; have\; to\; compute\; what\; will\; be\; the\; experience\; index\; of\; the\; ghost\; kind\; GXGX\; in\; the\; indefinite\; future,\; this\; is\; your\; task\; for\; today.}$

Note that when Tameem took the picture, $\mathrm{GXGX\; may\; already\; be\; greater\; than\; 00,\; because\; many\; ghosts\; may\; have\; scared\; one\; another\; at\; any\; moment\; between\; [-\infty ,T][-\infty ,T].}$

Input

The first line contains three integers $\mathrm{nn,\; aa\; and\; bb\; (1\le n\le 2000001\le n\le 200000,\; 1\le |a|\le 1091\le |a|\le 109,\; 0\le |b|\le 1090\le |b|\le 109) —\; the\; number\; of\; ghosts\; in\; the\; universe\; and\; the\; parameters\; of\; the\; straight\; line.}$

Each of the next $\mathrm{nn\; lines\; contains\; three\; integers\; xixi,\; VxiVxi,\; VyiVyi\; (-109\le xi\le 109-109\le xi\le 109,\; -109\le Vxi,Vyi\le 109-109\le Vxi,Vyi\le 109),\; where\; xixi\; is\; the\; current\; xx-coordinate\; of\; the\; ii-th\; ghost\; (and\; yi=a\cdot xi+byi=a\cdot xi+b).}$

It is guaranteed that no two ghosts share the same initial position, in other words, it is guaranteed that for all $(i,j)(i,j)\; xi\ne xjxi\ne xj\; for\; i\ne ji\ne j.$

Output

Output one line: experience index of the ghost kind $\mathrm{GXGX\; in\; the\; indefinite\; future.}$

Examples

Input

Copy

4 1 1
1 -1 -1
2 1 1
3 1 1
4 -1 -1

Output

Copy

8

Input

Copy

3 1 0
-1 1 0
0 0 -1
1 -1 -2

Output

Copy

6

Input

Copy

3 1 0
0 0 0
1 0 0
2 0 0

Output

Copy

0

Note

There are four collisions $(1,2,T-0.5)(1,2,T-0.5),\; (1,3,T-1)(1,3,T-1),\; (2,4,T+1)(2,4,T+1),\; (3,4,T+0.5)(3,4,T+0.5),\; where\; (u,v,t)(u,v,t)\; means\; a\; collision\; happened\; between\; ghosts\; uu\; and\; vv\; at\; moment\; tt.\; At\; each\; collision,\; each\; ghost\; gained\; one\; experience\; point,\; this\; means\; that\; GX=4\cdot 2=8GX=4\cdot 2=8.$

In the second test, all points will collide when $\mathrm{t=T+1t=T+1.}$

The red arrow represents the 1-st ghost velocity, orange represents the 2-nd ghost velocity, and blue represents the 3-rd ghost velocity.

题意：n只鬼当在同一时刻同一坐标遇见时会被对方吓到，问在时间为负无穷到正无穷范围（一开始没读清楚，到比赛结束第二题晚上才读清楚，可能是因为比赛结束后补D已经到一点头脑不清醒的缘故吧）内所有鬼被吓到次数的和。神奇的是，在T时刻，所有的鬼都在直线y=ax+b上，给你鬼在T时刻的横坐标和鬼的在xy轴的速度。

思路：本题求的其实就是鬼的行进路线所在直线间的交点数*2，易知两条直线在同一平面内只有平行和相交，由题意所说的T时刻所有的鬼在同一直线上知，鬼的行进路线如果平行只能是同时沿着T时刻的那条直线运动；而不平行的肯定会相交啦~我们先将题目中给的信息进行化简，如下图（字丑见谅）：

因此我们只要统计每个a*vx - vy(p)出现的次数即可，而对于在同一个a*vx-y的鬼(数量记为q，借助map进行储存，这个我是看了别人的代码才知道的==！），由于T时刻的x不同，因此，他们是不会相遇的。最后我们只要把每个鬼对应的p和q相减的值加起来即可。

代码实现如下：

 

#include <map>
#include <cstdio>
#include <cstdlib>
using namespace std;

const int maxn = 2e5 + 7;
typedef long long ll;
int n, a, b;
ll x[maxn], y[maxn];
map<ll,ll> mp1;
map<pair<ll, ll>, ll> mp2;

int main() {
    scanf("%d%d%d", &n, &a, &b);
    for(int i = 0; i < n; i++) {
        scanf("%*s%I64d%I64d", &x[i], &y[i]);
        mp1[a * x[i] - y[i]]++;
        mp2[make_pair(x[i], y[i])]++;
    }
    ll ans = 0;
    for(int i = 0; i < n; i++) {
        ans += mp1[a * x[i] - y[i]] - mp2[make_pair(x[i], y[i])];
    }
    printf("%I64d\n", ans);
    return 0;
}