Codeforces Round #222 (Div. 2)-mysql教程-PHP中文网

Codeforces Round #222 (Div. 2)

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发布： 2016-06-07 15:44:14

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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

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output

3 0 3

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input

2 4

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output

2 1 3

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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 <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 <= 6; i ++) {
        if (abs(a - i) < abs(b - i)) ans1++;
        if (abs(a - i) == abs(b - i)) ans2++;
        if (abs(a - i) > abs(b - i)) ans3++;
    }
    printf("%d %d %d\n", ans1, ans2, ans3);
    return 0;
}

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B题：随即1-k表示半决赛前k名直接晋级，剩下的人按时间排，贪心。

#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 < n; i ++)
        scanf("%d%d", &a[i], &b[i]);
}

void solve() {
    int l = 0, r = 0, i;
    for (i = 0; i < n / 2; i ++)
        an[i] = bn[i] = 1;
    for (i = 0; i < n; i ++) {
        if (a[l] < b[r]) {
            an[l++] = 1;
        }
        else {
            bn[r++] = 1;
        }
    }
    for (i = 0; i < n; i ++)
        printf("%d", an[i]);
    printf("\n");
    for (i = 0; i < n; i ++)
        printf("%d", bn[i]);
    printf("\n");
}

int main() {
    init();
    solve();
    return 0;
}

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C题：给定k步，要求填到只剩一块连接的空白。搜索题

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
using namespace std;

const int N = 505;
const int d[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

char g[N][N];
int n, m, k, pn, sum, Max_v, Max, snum;
struct P {
    int x, y, v;
} p[N * N];

int cmp(P a, P b) {
    return a.v > 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 < n; i ++)
        scanf("%s", g[i]);
}

void dfs1(int x, int y) {
    g[x][y] = 'X'; p[pn].v++;
    for (int i = 0; i < 4; i ++) {
        int xx = x + d[i][0];
        int yy = y + d[i][1];
        if (xx >= 0 && xx < n && yy >= 0 && yy < m && g[xx][yy] == '.')
            dfs1(xx, yy);
    }
}

void dfs2(int x, int y) {
    if (snum == sum - k) {
        return;
    }
    g[x][y] = '.'; snum ++;
    for (int i = 0; i < 4; i ++) {
        int xx = x + d[i][0];
        int yy = y + d[i][1];
        if (xx >= 0 && xx < n && yy >= 0 && yy < m && g[xx][yy] == 'X')
            dfs2(xx, yy);
    }
}

void print() {
    for (int i = 0; i < n; i ++)
        printf("%s\n", g[i]);
}

void solve() {
    for (int i = 0; i < n; i ++)
        for (int j = 0; j < m; j ++) {
            if (g[i][j] == '.') {
                dfs1(i, j);
                sum += p[pn].v;
                if (Max < p[pn].v) {
                    Max_v = pn;
                    Max = p[pn].v;
                }
                p[pn].x = i; p[pn].y = j; pn++;
            }
        }
    dfs2(p[Max_v].x, p[Max_v].y);
    print();
}

int main() {
    init();
    solve();
    return 0;
}

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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 <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 < (S a, S b) {
        return a.c > 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 < b.a;
}

void init() {
    int i;
    scanf("%d%d%d", &n, &m, &s);
    for (i = 0; i < m; i ++) {
        scanf("%d", &bd[i].a);
        bd[i].id = i;
    }
    for (i = 0; i < n; i ++) {
        scanf("%d", &st[i].b);
        st[i].id = i;
    }
    for (i = 0; i < n; i ++)
        scanf("%d", &st[i].c);
    sort(bd, bd + m, cmp1);
    sort(st, st + n, cmp);
}

bool judge1(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 < t.c) return false;
        ss -= t.c;
        int e = i - time + 1;
        if (e < 0) e = 0;
        for (int j = i; j >= 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 < t.c) return false;
        ss -= t.c;
    }
    return true;
}

void solve() {
    int l = 0, r = m;
    if (!judge(r)) {
        printf("NO\n"); return;
    }
    while (l < r) {
        int mid = (l + r) / 2;
        if (judge(mid)) r = mid;
        else l = mid + 1;
    }
    judge1(r);
    printf("YES\n");
    for (int i = 0; i < m - 1; i++)
        printf("%d ", ans[i] + 1);
    printf("%d\n", ans[m - 1] + 1);
}

int main() {
    init();
    solve();
    return 0;
}

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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，

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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 <stdio.h>
#include <string.h>
#include <algorithm>
#define min(a,b) (a)<(b)?(a):(b)
#define max(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 < n; i++)
        scanf("%d", &s[i]);
    scanf("%d%*c", &m);
    for (int j = 0; j < m; j++)
        scanf("%c%*c%d%*c", &c[j], &t[j]);
}

int DP(int state, int num) {
    if (dp[state] != INF) return dp[state];
    int &ans = dp[state];
	ans = 0;
    if (c[num] == 'p') {
        int bit;
        for (bit = 0; bit < m; bit++)
            if ((state & (1<<bit)))
                break;
        if (t[num] == 1)
            ans = (DP((state^(1<<bit)), num + 1) + s[bit]);
        else
            ans = (DP((state^(1<<bit)), num + 1) - s[bit]);
    }
    else if (c[num] == 'b') {
		if (t[num] == 1) ans = -INF;
		else ans = INF;
        for (int i = 0; i < m; i++) {
            if ((state & (1<<i))) {
				if (t[num] == 1)
					ans = max(ans, DP((state^(1<<i)), num + 1));
				else 
					ans = min(ans, DP((state^(1<<i)), num + 1));
            }
        }
    }
    return ans;
}

void solve() {
    sort(s, s + n, cmp);
    st = (1<<m) - 1;
    printf("%d\n", DP(st, 0));
}

int main() {
    init();
    solve();
    return 0;
}

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