Codeforces Round #222 (Div. 2)
Jun 07, 2016 pm 03:44 PMA. 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(shuí)小就誰(shuí)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>

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