Codeforces Round #222 (Div. 2)

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

memory limit per test

input

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题：a，bi两个数字，扔一个色字，求分别与a，b求差的绝对值，谁小就谁赢，相等平局，输出情况。

水：

#include 
#include 
#include 
#include 

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;
}

#include 
#include 
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 

C题：给定k步，要求填到只剩一块连接的空白。搜索题


#include 
#include 
#include 
#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 

D题：m个bug每个bug有级别，n个人，每个人有级别和需求，现在总共有s个需求，求最少天数完成的方法，并且输出方案。

思路：二分+贪心+优先队列优化

#include 
#include 
#include 
#include 
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_queueQ;
    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题:dota2 进行 bp操作，每个英雄有一个能力值，玩家1,2分别进行b,p操作，每个玩家都尽量往好了取，要求最后能力值的差，
思路：dp+贪心+位运算，对于一个玩家进行pick时，肯定选能力值最大的，这是贪心。进行ban时。要把所有情况找出来。用dp的记忆化搜索。对于状态利用2进制的位运算。
代码：
#include 
#include 
#include 
#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