Codeforces Round #222 (Div. 2)

A. Playing with Dicetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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 numberA, The second player wrote numberB. 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 integersAAndB(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 numberAIs closer to numberXThan numberB, If |A? -?X|? B? -?X|.

Question A: a and bi are two numbers, throwing a color word and finding the absolute values of the difference with A and B respectively. If the number is small, the person wins, the draw is equal, and the result is output.

Water:

#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 <= 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;}
    
   
  
 

Question B: 1-k immediately indicates that the first k of the semi-finals will be promoted directly, and the rest will be ranked by time, greedy.

#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 < 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;}
  
 

Question C: Given k steps, it is required to fill in only one blank link. Search questions

#include 
 
  #include 
  
   #include #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: m bugs. Each bug has a level of n people. Each person has a level and a requirement. Now there are a total of s requirements. Find a method that can be completed in at least a few days and output a solution.

Idea: Binary + greedy + priority queue Optimization

#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 < (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
    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_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 < 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;}
   
  
 

Question E: dota2 performs bp operations. Each hero has a skill value. players perform B and p Operations respectively. Each player tries its best to obtain the skill value, which requires a poor final skill value,

Idea: dp + greedy + bit operations. When a player performs a pick operation, the maximum value is certainly selected. This is greedy. When ban is performed. Find out all the situations. Use the memory of dp to search. The bitwise operation of the status is in hexadecimal notation.

Code:

#include 
   
    #include 
    
     #include #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<