A. Playing with Dice time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown.

A. Playing with Dice

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.

The first player wrote number?a, the second player wrote number?b. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?

Input

The single line contains two integers?a?and?b?(1?≤?a,?b?≤?6)?— the numbers written on the paper by the first and second player, correspondingly.

Output

Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.

Sample test(s)

input

2 5

output

3 0 3

input

2 4

output

2 1 3

Note

The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.

You can assume that number?a?is closer to number?x?than number?b, if?|a?-?x|?|b?-?x|.

A題：a，bi兩個(gè)數(shù)字，扔一個(gè)色字，求分別與a，b求差的絕對(duì)值，誰小就誰贏，相等平局，輸出情況。

水：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int a, b;

int main() {
    int ans1 = 0, ans2 = 0, ans3 = 0;
    scanf("%d%d", &a, &b);
    for (int i = 1; i  abs(b - i)) ans3++;
    }
    printf("%d %d %d\n", ans1, ans2, ans3);
    return 0;
}</string.h></math.h></stdlib.h></stdio.h>

B題：隨即1-k表示半決賽前k名直接晉級(jí)，剩下的人按時(shí)間排，貪心。

#include <stdio.h>
#include <string.h>
const int N = 100005;
int n, a[N], b[N];
int an[N], bn[N];

void init() {
    scanf("%d", &n);
    memset(an, 0, sizeof(an));
    memset(bn, 0, sizeof(bn));
    for (int i = 0; i <br>
C題：給定k步，要求填到只剩一塊連接的空白。搜索題


<pre class="brush:php;toolbar:false">#include <stdio.h>
#include <string.h>
#include <algorithm>
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a) b.v;
}

void init() {
    sum = 0; Max = 0; snum = 0;
    memset(p, 0, sizeof(p));
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i = 0 && xx = 0 && yy = 0 && xx = 0 && yy <br>
D題：m個(gè)bug每個(gè)bug有級(jí)別，n個(gè)人，每個(gè)人有級(jí)別和需求，現(xiàn)在總共有s個(gè)需求，求最少天數(shù)完成的方法，并且輸出方案。

<p>思路：二分+貪心+優(yōu)先隊(duì)列優(yōu)化</p>

<pre class="brush:php;toolbar:false">#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;

const int N = 100005;

int n, m, s, a[N], ans[N];

struct S {
    int b, c, id;
    friend bool operator  b.c;
    }
} st[N];

struct B {
    int a, id;
} bd[N];

int cmp(S a, S b) {
    return a.b > b.b;
}

int cmp1(B a, B b) {
    return a.a Q;
    for (int i = m - 1; i >= 0; i -= time) {
        while (st[sn].b >= bd[i].a && sn != n) {Q.push(st[sn++]);}
        if (Q.empty()) return false;
        S t = Q.top(); Q.pop();
        if (ss = e; j--) {
            ans[bd[j].id] = t.id;
        }
    }
    return true;
}

bool judge(int time) {
    int ss = s, sn = 0;
    priority_queue<s>Q;
    for (int i = m - 1; i >= 0; i -= time) {
        while (st[sn].b >= bd[i].a && sn != n) {Q.push(st[sn++]);}
        if (Q.empty()) return false;
        S t = Q.top(); Q.pop();
        if (ss <br>
</s><p>E題:dota2 進(jìn)行 bp操作，每個(gè)英雄有一個(gè)能力值，玩家1,2分別進(jìn)行b,p操作，每個(gè)玩家都盡量往好了取，要求最后能力值的差，</p>
<p>思路：dp+貪心+位運(yùn)算，對(duì)于一個(gè)玩家進(jìn)行pick時(shí)，肯定選能力值最大的，這是貪心。進(jìn)行ban時(shí)。要把所有情況找出來。用dp的記憶化搜索。對(duì)于狀態(tài)利用2進(jìn)制的位運(yùn)算。</p>
<p>代碼：</p>
<pre class="brush:php;toolbar:false">#include <stdio.h>
#include <string.h>
#include <algorithm>
#define min(a,b) (a)(b)?(a):(b)
using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 1111111;
const int N = 105;
const int M = 25;

int cmp(int a, int b) {
    return a > b;
}

int n, m, s[N], c[M], t[M], dp[MAXN], st;

void init() {
    memset(dp, INF, sizeof(dp));
    scanf("%d", &n);
    for (int i = 0; i <br>
<br>



</algorithm></string.h></stdio.h>