Codeforces 258 Div2

时间：2014-07-26 09:52:57 阅读：284 评论：0 收藏：0 [点我收藏+]

标签：style blog http color os io 2014 for

A题，n*m根木棍，相交放置，轮流取走相交的两根，最后谁不能行动，则输掉。

min(n,m)&1 为1则先取者赢。

B题，给定一个长度为n，且各不相同的数组，问能否通过交换连续一段L....R使得变成单调递增。

如果一开始就是递增的，那么直接输出L。。。R就是1 1，交换一个就行了；否则判断中间是否有且一段单调递减，且两端交换后使得数组递增。

代码：

  1 //Template updates date: 20140718
  2 #include <iostream>
  3 #include <sstream>
  4 #include <cstdio>
  5 #include <climits>
  6 #include <ctime>
  7 #include <cctype>
  8 #include <cstring>
  9 #include <cstdlib>
 10 #include <string>
 11 #include <stack>
 12 #include <set>
 13 #include <map>
 14 #include <cmath>
 15 #include <vector>
 16 #include <queue>
 17 #include <algorithm>
 18 #define  esp 1e-6
 19 #define  inf 0x3f3f3f3f
 20 #define  pi acos(-1.0)
 21 #define  pb push_back
 22 #define  lson l, m, rt<<1
 23 #define  rson m+1, r, rt<<1|1
 24 #define  lowbit(x) (x&(-x))
 25 #define  mp(a, b) make_pair((a), (b))
 26 #define  bit(k) (1<<(k))
 27 #define  iin  freopen("pow.in", "r", stdin);
 28 #define  oout freopen("pow.out", "w", stdout);
 29 #define  in  freopen("solve_in.txt", "r", stdin);
 30 #define  out freopen("solve_out.txt", "w", stdout);
 31 #define  bug puts("********))))))");
 32 #define  Inout iin oout
 33 #define  inout in out
 34 
 35 #define  SET(a, v) memset(a, (v), sizeof(a))
 36 #define  SORT(a)   sort((a).begin(), (a).end())
 37 #define  REV(a)    reverse((a).begin(), (a).end())
 38 #define  READ(a, n) {REP(i, n) cin>>(a)[i];}
 39 #define  REP(i, n) for(int i = 0; i < (n); i++)
 40 #define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
 41 #define  Rep(i, base, n) for(int i = (base); i < (n); i++)
 42 #define  REPS(s, i) for(int i = 0; (s)[i]; i++)
 43 #define  pf(x) ((x)*(x))
 44 #define  mod(n) ((n))
 45 #define  Log(a, b) (log((double)b)/log((double)a))
 46 #define Srand() srand((int)time(0))
 47 #define random(number) (rand()%number)
 48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
 49 
 50 using namespace std;
 51 typedef long long  LL;
 52 typedef unsigned long long ULL;
 53 typedef vector<int> VI;
 54 typedef pair<int,int> PII;
 55 typedef vector<PII> VII;
 56 typedef vector<PII, int> VIII;
 57 typedef VI:: iterator IT;
 58 typedef map<string, int> Mps;
 59 typedef map<int, int> Mpi;
 60 typedef map<int, PII> Mpii;
 61 typedef map<PII, int> Mpiii;
 62 
 63 const int maxn = 100000 + 100;
 64 int a[maxn];
 65 
 66 int main() {
 67 
 68     int n;
 69     Rep(i, scanf("%d", &n), n+1) scanf("%d", a+i);
 70     int l = 0, r = 0;
 71     int ok = 0;
 72     a[n+1] = inf;
 73     Rep(i, 1, n+2) {
 74         if(a[i] > a[i-1]) {
 75             if(!l)
 76                 continue;
 77             else {
 78                 if(ok && !r)
 79                     r = i-1;
 80             }
 81         } else if(a[i] < a[i-1]) {
 82             if(!ok)
 83                 ok = 1, l = i-1;
 84             if(r) {
 85                 cout<<"no"<<endl;
 86                 return 0;
 87             }
 88         } else {
 89             cout<<"no"<<endl;
 90             return 0;
 91         }
 92     }
 93     if(l) {
 94         if(a[l] < a[r+1] && a[r] > a[l-1]) {
 95             cout<<"yes"<<endl;
 96             cout<<l <<‘ ‘<<r<<endl;
 97         } else {
 98             cout<<"no"<<endl;
 99         }
100     } else {
101         cout<<"yes"<<endl;
102         cout<<1<<‘ ‘<<1<<endl;
103     }
104     return 0;
105 }

View Code

写得挫。

C题，3个球队，总共n场比赛，已进行k场，知道两个球队之间赢得场数的绝对值，如|x1-x2| = d1, |x2-x3| = d2, 问剩下的比赛中是否存在一种可能使得三个队伍赢得场数相等。

分析：题意知道每场比赛只用知道谁赢就行，不用顾忌各种规则，前k场中分别可能情况：

令x = x2;

x+d1, x, x+d2;

x+d1, x, x-d2;

x-d1, x, x+d2;

x-d1, x, x-d1;

对应4种情况求出x，在求出x1,x2,x3,判断满足0<=xi<=n/3 && xi <= k;

就可以了。

代码：

 1 //Template updates date: 20140718
 2 #include <iostream>
 3 #include <sstream>
 4 #include <cstdio>
 5 #include <climits>
 6 #include <ctime>
 7 #include <cctype>
 8 #include <cstring>
 9 #include <cstdlib>
10 #include <string>
11 #include <stack>
12 #include <set>
13 #include <map>
14 #include <cmath>
15 #include <vector>
16 #include <queue>
17 #include <algorithm>
18 #define  esp 1e-6
19 #define  inf 0x3f3f3f3f
20 #define  pi acos(-1.0)
21 #define  pb push_back
22 #define  lson l, m, rt<<1
23 #define  rson m+1, r, rt<<1|1
24 #define  lowbit(x) (x&(-x))
25 #define  mp(a, b) make_pair((a), (b))
26 #define  bit(k) (1<<(k))
27 #define  iin  freopen("pow.in", "r", stdin);
28 #define  oout freopen("pow.out", "w", stdout);
29 #define  in  freopen("solve_in.txt", "r", stdin);
30 #define  out freopen("solve_out.txt", "w", stdout);
31 #define  bug puts("********))))))");
32 #define  Inout iin oout
33 #define  inout in out
34 
35 #define  SET(a, v) memset(a, (v), sizeof(a))
36 #define  SORT(a)   sort((a).begin(), (a).end())
37 #define  REV(a)    reverse((a).begin(), (a).end())
38 #define  READ(a, n) {REP(i, n) cin>>(a)[i];}
39 #define  REP(i, n) for(int i = 0; i < (n); i++)
40 #define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
41 #define  Rep(i, base, n) for(int i = (base); i < (n); i++)
42 #define  REPS(s, i) for(int i = 0; (s)[i]; i++)
43 #define  pf(x) ((x)*(x))
44 #define  mod(n) ((n))
45 #define  Log(a, b) (log((double)b)/log((double)a))
46 #define Srand() srand((int)time(0))
47 #define random(number) (rand()%number)
48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
49 
50 using namespace std;
51 typedef long long  LL;
52 typedef unsigned long long ULL;
53 typedef vector<int> VI;
54 typedef pair<int,int> PII;
55 typedef vector<PII> VII;
56 typedef vector<PII, int> VIII;
57 typedef VI:: iterator IT;
58 typedef map<string, int> Mps;
59 typedef map<int, int> Mpi;
60 typedef map<int, PII> Mpii;
61 typedef map<PII, int> Mpiii;
62 
63 LL n, k, d1, d2;
64 int a[][2] = {{-1, 1} ,{-1, -1}, {1, 1}, {1, -1}};
65 
66 int main() {
67     int T;
68     for(int t = scanf("%d", &T); t <= T; t++) {
69         scanf("%I64d%I64d%I64d%I64d", &n, &k, &d1, &d2);
70         if(n%3) {
71             puts("no");
72             continue;
73         }
74         int ok = 0;
75         LL nn;
76         REP(i, 4) {
77             nn = k;
78             LL tmp = (LL)a[i][0]*d1+(LL)a[i][1]*d2;
79             nn -= tmp;
80             if(nn < 0 || nn%3)
81                 continue;
82             nn /= 3;
83             if(nn > n/3 || nn > k)
84                 continue;
85             LL x1 = nn+(LL)a[i][0]*d1;
86             LL x2 = nn+(LL)a[i][1]*d2;
87             if(x1 < 0 || x1 > n/3 || x2 < 0 || x2 > n/3 || x1 > k || x2 > k)
88                 continue;
89             ok = 1;
90             break;
91         }
92         puts(ok ? "yes" : "no");
93     }
94     return 0;
95 }

View Code

D题，给定仅有‘a‘，‘a‘组成的字符串，定义一种子串：通过合并连续的字母最后可以得到一个回文串。如”aaabbaaa"最后得到"aba"。

很容易想到一个字串首尾字符相等时就为这样的要求的特殊字串。

题目要求奇数长度和偶数长度这样的字串个数，统计每个字符分别用dp[0][s[i]-‘a‘], dp[1][s[i]-‘a‘],表示位于i及之前的奇数位置和偶数位置的s[i]字符个数，

这样偶数长度ans0 += dp[!(i&1)][s[i]-‘a‘], 奇数长度ans1 += dp[i&1][s[i]-‘a‘];

代码：

 1 //Template updates date: 20140718
 2 #include <iostream>
 3 #include <sstream>
 4 #include <cstdio>
 5 #include <climits>
 6 #include <ctime>
 7 #include <cctype>
 8 #include <cstring>
 9 #include <cstdlib>
10 #include <string>
11 #include <stack>
12 #include <set>
13 #include <map>
14 #include <cmath>
15 #include <vector>
16 #include <queue>
17 #include <algorithm>
18 #define  esp 1e-6
19 #define  inf 0x3f3f3f3f
20 #define  pi acos(-1.0)
21 #define  pb push_back
22 #define  lson l, m, rt<<1
23 #define  rson m+1, r, rt<<1|1
24 #define  lowbit(x) (x&(-x))
25 #define  mp(a, b) make_pair((a), (b))
26 #define  bit(k) (1<<(k))
27 #define  iin  freopen("pow.in", "r", stdin);
28 #define  oout freopen("pow.out", "w", stdout);
29 #define  in  freopen("solve_in.txt", "r", stdin);
30 #define  out freopen("solve_out.txt", "w", stdout);
31 #define  bug puts("********))))))");
32 #define  Inout iin oout
33 #define  inout in out
34 
35 #define  SET(a, v) memset(a, (v), sizeof(a))
36 #define  SORT(a)   sort((a).begin(), (a).end())
37 #define  REV(a)    reverse((a).begin(), (a).end())
38 #define  READ(a, n) {REP(i, n) cin>>(a)[i];}
39 #define  REP(i, n) for(int i = 0; i < (n); i++)
40 #define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
41 #define  Rep(i, base, n) for(int i = (base); i < (n); i++)
42 #define  REPS(s, i) for(int i = 0; (s)[i]; i++)
43 #define  pf(x) ((x)*(x))
44 #define  mod(n) ((n))
45 #define  Log(a, b) (log((double)b)/log((double)a))
46 #define Srand() srand((int)time(0))
47 #define random(number) (rand()%number)
48 #define random_range(a, b) (int)(((double)rand()/RAND_MAX)*(b-a) + a)
49 
50 using namespace std;
51 typedef long long  LL;
52 typedef unsigned long long ULL;
53 typedef vector<int> VI;
54 typedef pair<int,int> PII;
55 typedef vector<PII> VII;
56 typedef vector<PII, int> VIII;
57 typedef VI:: iterator IT;
58 typedef map<string, int> Mps;
59 typedef map<int, int> Mpi;
60 typedef map<int, PII> Mpii;
61 typedef map<PII, int> Mpiii;
62 
63 
64 const int maxn = 100000 + 100;
65 char s[maxn];
66 LL dp[2][2];
67 int main() {
68     
69     scanf("%s", s);
70     LL ans1 = 0, ans2 = 0;
71     REPS(s, i) {
72         int x = s[i]-‘a‘;
73         int y = i&1;
74         dp[y][x]++;
75         ans1+=dp[y][x];
76         ans2+=dp[y^1][x];
77     }
78     cout<<ans2<<‘ ‘<<ans1<<endl;
79     return 0;
80 }