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. 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 <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;}

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;}

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;}

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;}

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 <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;}