2015 Multi-University Training Contest 2

hdu 5301  Buildings

机智题，卡了好久，最后小坤纸上去敲过了。。做法：假设n < m，定义ans = (n + 1) / 2，ret = min(b, m - b + 1)。。当ret <= ans，对于当黑格子处于左上边的情况，就可以打横来覆盖左边的部分。当ret > ans的时候，就没办法打横覆盖了，ans = min(max(n - a, a - 1), ret)（注意a-1 != n-a）。最后还要特判一下黑格子处于中心的情况。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<string>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<set>
 8 using namespace std;
 9 
10 
11 #define inf 0x3f3f3f3f
12 #define eps 1e-9
13 #define FOR(i,s,t) for(int i = s; i < t; ++i )
14 #define REP(i,s,t) for( int i = s; i <= t; ++i )
15 #define pii pair<int,int>
16 #define MP make_pair
17 #define lson id << 1 , l , mid
18 #define rson id << 1 | 1 , mid + 1 , r
19 #define LL long long
20 #define ULL unsigned long long
21 #define N ( 500000 + 10  )
22 #define M ( 1000000 + 10)
23 #define mod 1000000007
24 
25 int main(){
26     int a, n, m, b;
27     while(scanf("%d%d%d%d", &n, &m, &a, &b) != EOF){
28         if(m <= 2 || n <= 2){
29             puts("1"); continue;
30         }
31         if(n % 2 && m % 2 && a == (n + 1) / 2 && b == (m + 1) / 2){
32             if(n == m) printf("%d\n", a - 1);
33             else printf("%d\n", min(b, a));
34             continue;
35         }
36         if(n > m)
37             swap(n, m), swap(a, b);
38         int ans = (n + 1) / 2, ret = min(b, m - b + 1);
39         if(ret > ans && a - 1 != n - a)
40             ans = min(max(a - 1, n - a), ret);
41         printf("%d\n", ans);
42     }
43     return 0;
44 }
View Code

hdu 5302  Connect the Graph

构造题。当n >= 5时肯定有解。n为奇数，假设b2 = n, w2 = n，直接可以构造出1，2，3……n一个环满足b2 = n，构造1，3，5……2，4……1这样的环满足w2 = n。n为偶数，假设b2 = n, w2 = n，可以构造出1，2，3……n满足b2 = n，构造1，3，5……2，n，n-2，n-4……4满足w2 = n。对于b2 != n, w2 != n直接拆环就好了。对于n <= 4爆搜一下就好了。（实际上b0, b1, b2, w0, w1, w2都大于等于1，小于5的情况好像只有1,2,1,1,2,1才有解，其他无解，不需要搜）

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<string>
  5 #include<cstring>
  6 #include<algorithm>
  7 #include<set>
  8 using namespace std;
  9 
 10 
 11 #define inf 0x3f3f3f3f
 12 #define eps 1e-6
 13 #define MP make_pair
 14 #define LL long long
 15 #define N 1000010
 16 #define M 200020
 17 #define mod 1000000007
 18 
 19 int n, a[2][3], c[N];
 20 void work(int col){
 21     if(a[col][2] == n){
 22         for(int i = 1; i < n; ++i)
 23             printf("%d %d %d\n", c[i], c[i+1], col);
 24         printf("%d %d %d\n", c[n], c[1], col);
 25     }
 26     else{
 27         for(int i = 1; i <= a[col][2] + 1; ++i)
 28             printf("%d %d %d\n", c[i], c[i+1], col);
 29         for(int i = 3; i <= a[col][1]; i += 2)
 30             printf("%d %d %d\n", c[a[col][2]+i], c[a[col][2]+i+1], col);
 31     }
 32 }
 33 bool ok;
 34 int cnt, uu[10], vv[10], col[10], use[5][2], ww[3], bb[3];
 35 void dfs(int cur){
 36     if(ok) return ;
 37     if(cur == cnt){
 38         memset(use, 0, sizeof use);
 39         ww[0] = ww[1] = ww[2] = bb[0] = bb[1] = bb[2] = 0;
 40         for(int i = 0; i < cnt; ++i){
 41             if(col[i] == 0)
 42                 use[uu[i]][0]++, use[vv[i]][0]++;
 43             if(col[i] == 1)
 44                 use[uu[i]][1]++, use[vv[i]][1]++;
 45         }
 46         for(int i = 1; i <= n; ++i){
 47             if(use[i][0] > 2 || use[i][1] > 2) return ;
 48             ww[use[i][0]]++, bb[use[i][1]]++;
 49         }
 50         for(int i = 0; i < 3; ++i)
 51             if(a[0][i] != ww[i] || a[1][i] != bb[i]) return ;
 52         ok = 1;
 53         printf("%d\n", a[0][1]/2 + a[0][2] + a[1][1]/2 + a[1][2]);
 54         for(int i = 0; i < cnt; ++i)
 55             if(col[i] == 0 || col[i] == 1)
 56                 printf("%d %d %d\n", uu[i], vv[i], col[i]);
 57         return ;
 58     }
 59     col[cur] = 0;
 60     dfs(cur + 1);
 61     col[cur] = 1;
 62     dfs(cur + 1);
 63     col[cur] = 2;
 64     dfs(cur + 1);
 65 }
 66 void gao(){
 67     cnt = 0;
 68     ok = 0;
 69     for(int i = 1; i <= n; ++i)
 70         for(int j = i + 1; j <= n; ++j)
 71             uu[cnt] = i, vv[cnt] = j, col[cnt++] = 0;
 72     dfs(0);
 73     if(!ok) puts("-1");
 74 }
 75 int main(){
 76     //freopen("tt.txt", "r", stdin);
 77     int cas;
 78     scanf("%d", &cas);
 79     while(cas--){
 80         scanf("%d%d%d%d%d%d", &a[0][0], &a[0][1], &a[0][2], &a[1][0], &a[1][1], &a[1][2]);
 81         if(a[0][1] % 2 || a[1][1] % 2){
 82             puts("-1"); continue;
 83         }
 84         n = a[0][0] + a[0][1] + a[0][2];
 85         if(n <= 4){
 86             gao();
 87             continue;
 88         }
 89         printf("%d\n", a[0][1]/2 + a[0][2] + a[1][1]/2 + a[1][2]);
 90         if(n % 2){
 91             for(int i = 1; i <= n; ++i)
 92                 c[i] = i;
 93             work(0);
 94             int cnt = 0;
 95             for(int i = 1; i <= n; i += 2)
 96                 c[++cnt] = i;
 97             for(int i = 2; i <= n; i += 2)
 98                 c[++cnt] = i;
 99             work(1);
100         }
101         else{
102             for(int i = 1; i <= n; ++i)
103                 c[i] = i;
104             work(0);
105             int cnt = 0;
106             for(int i = 1; i <= n; i += 2)
107                 c[++cnt] = i;
108             c[++cnt] = 2;
109             for(int i = n; i >= 4; i -= 2)
110                 c[++cnt] = i;
111             work(1);
112         }
113     }
114     return 0;
115 }
View Code

 hdu 5303  Delicious Apples

一道十分好的贪心题。最多只能绕一次环。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<string>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<set>
 8 using namespace std;
 9 
10 
11 #define inf 0x3f3f3f3f
12 #define eps 1e-9
13 #define FOR(i,s,t) for(int i = s; i < t; ++i )
14 #define REP(i,s,t) for( int i = s; i <= t; ++i )
15 #define pii pair<int,int>
16 #define MP make_pair
17 #define lson id << 1 , l , mid
18 #define rson id << 1 | 1 , mid + 1 , r
19 #define LL long long
20 #define ULL unsigned long long
21 #define N 1000010
22 #define M 200020
23 #define mod 1000000007
24 
25 int a[N], a1[N], a2[N];
26 LL p1[N], p2[N];
27 int main(){
28     int cas;
29     scanf("%d", &cas);
30     while(cas--){
31         int L, n, k;
32         scanf("%d%d%d", &L, &n, &k);
33         int s = 0, mid = L / 2;
34         for(int i = 0; i < n; ++i){
35             int pos, num;
36             scanf("%d%d", &pos, &num);
37             for(int j = 0; j < num; ++j)
38                 a[++s] = pos;
39         }
40         k = min(s, k);
41         int c1 = 0, c2 = 0;
42         for(int i = 1; i <= s; ++i){
43             if(a[i] <= mid) a1[++c1] = a[i];
44             else a2[++c2] = L - a[i];
45         }
46         sort(a1 + 1, a1 + 1 + c1);
47         sort(a2 + 1, a2 + 1 + c2);
48         for(int i = 1; i <= c1; ++i){
49             if(i <= k) p1[i] = a1[i];
50             else p1[i] = p1[i-k] + a1[i];
51         }
52         for(int i = 1; i <= c2; ++i){
53             if(i <= k) p2[i] = a2[i];
54             else p2[i] = p2[i-k] + a2[i];
55         }
56         LL ans = (p1[c1] + p2[c2]) * 2;
57         for(int i = 1; i <= c1 && i <= k; ++i){
58             int lef = c1 - i, rig = c2 - (k - i);
59             ans = min(ans, (p1[lef] + p2[rig]) * 2 + L);
60         }
61         printf("%I64d\n", ans);
62     }
63     return 0;
64 }
View Code

hdu 5307  He is Flying

一道FFT的题目。用复数模板的FFT会因为精度误差而wa，又找了一个用快速数论变换的模板。中间有一个O(1)longlong内的乘法，我也不知道为什么这样算是对的。。

设所有的答案为 sigma(j - i +1) * X^(Sj - Si)，拓展开来就可以得到题解的式子，可以用FFT求出所有正数的答案。ans0再另外处理一下。

（c++交上去是超时的，g++才能过）

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <string>
  5 #include <cstring>
  6 #include <algorithm>
  7 #include <set>
  8 using namespace std;
  9 
 10 
 11 #define inf 0x3f3f3f3f
 12 #define eps 1e-5
 13 #define pii pair<int,int>
 14 #define MP make_pair
 15 #define LL long long
 16 #define N 100010
 17 #define M 200020
 18 #define mod 1000000007
 19 
 20 int read(){
 21     int ret = 0; char ch = getchar();
 22     while(ch < '0' || ch > '9') ch = getchar();
 23     while(ch >= '0' && ch <= '9') ret = ret * 10 + ch - '0', ch = getchar();
 24     return ret;
 25 }
 26 void out(LL x){
 27     if(x > 9) out(x / 10);
 28     putchar(x % 10 + '0');
 29 }
 30 const LL P = 50000000001507329LL, NN = 1LL<<18, G = 3;
 31 
 32 LL Mul(LL x, LL y){
 33     return (x * y - (LL)(x / (long double)P * y + 1e-3) * P + P) % P;
 34 }
 35 LL qpow(LL x, LL k, LL m){
 36     x %= m, k %= m;
 37     LL ret = 1;
 38     while(k){
 39         if(k & 1) ret = Mul(ret, x);
 40         x = Mul(x, x);
 41         k >>= 1;
 42     }
 43     return ret;
 44 }
 45 LL wn[25];
 46 void getWn(){
 47     for(int i = 0; i <= 20; ++i){
 48         int t = 1 << i;
 49         wn[i] = qpow(G, (P - 1) / t, P);
 50     }
 51 }
 52 void Rader(LL *a, int len){
 53     int k;
 54     for(int i = 1, j = len / 2; i < len - 1; ++i){
 55         if(i < j) swap(a[i], a[j]);
 56         k = len / 2;
 57         while(j >= k) j -= k, k /= 2;
 58         if(j < k) j += k;
 59     }
 60 }
 61 void FFT(LL *a, int len, int on){
 62     Rader(a, len);
 63     int id = 0;
 64     for(int h = 2; h <= len; h <<= 1){
 65         id++;
 66         for(int j = 0; j < len; j += h){
 67             LL w = 1;
 68             for(int k = j; k < j + h / 2; ++k){
 69                 LL u = a[k];
 70                 LL t = Mul(a[k+h/2], w);
 71                 a[k] = u + t;
 72                 if(a[k] >= P) a[k] -= P;
 73                 a[k+h/2] = u - t + P;
 74                 if(a[k+h/2] >= P) a[k+h/2] -= P;
 75                 w = Mul(w, wn[id]);
 76             }
 77         }
 78     }
 79     if(on == -1){
 80         for(int i = 1; i < len / 2; ++i)
 81             swap(a[i], a[len-i]);
 82         LL inv = qpow(len, P-2, P);
 83         for(int i = 0; i < len; ++i)
 84             a[i] = Mul(a[i], inv);
 85     }
 86 }
 87 void Conv(LL *a, LL *b, int n){
 88     FFT(a, n, 1);
 89     FFT(b, n, 1);
 90     for(int i = 0; i < n; ++i)
 91         a[i] = Mul(a[i], b[i]);
 92     FFT(a, n, -1);
 93 }
 94 LL s[N], p[N], a[M], b[M], ans[M];
 95 int main(){
 96     getWn();
 97     int cas;
 98     scanf("%d", &cas);
 99     for(int i = 1; i <= N / 2; ++i)
100         p[i] = p[i-1] + (LL)(i + 1) * (LL)i / 2;
101     while(cas--){
102         int n;
103         scanf("%d", &n);
104         for(int i = 1; i <= n; ++i){
105             int x;
106             x = read();
107             s[i] = s[i-1] + x;
108         }
109         LL ans0 = 0;
110         int pre = s[0], cnt = 0;
111         s[n+1] = -1;
112         for(int i = 1; i <= n + 1; ++i){
113             if(s[i] == pre)
114                 cnt++;
115             else
116                 ans0 += p[cnt], cnt = 0, pre = s[i];
117         }
118         int len = 1;
119         while(len <= s[n] * 2) len <<= 1;
120         memset(a, 0, sizeof a);
121         memset(b, 0, sizeof b);
122         for(int i = 1; i <= n; ++i)
123             a[s[i]] += i, b[s[n] - s[i-1]]++;
124         Conv(a, b, len);
125         for(int i = 0; i < len; ++i)
126             ans[i] = a[i];
127         memset(a, 0, sizeof a);
128         memset(b, 0, sizeof b);
129         for(int i = 1; i <= n; ++i)
130             a[s[i]]++, b[s[n] - s[i-1]] += i-1;
131         Conv(a, b, len);
132         for(int i = 0; i < len; ++i){
133             ans[i] = ans[i] - a[i];
134             if(ans[i] < 0) ans[i] += P;
135             if(ans[i] >= P) ans[i] -= P;
136         }
137         printf("%I64d\n", ans0);
138         for(int i = 1; i <= s[n]; ++i)
139             out(ans[i+s[n]]), puts("");
140     }
141     return 0;
142 }
View Code

hdu 5308  I Wanna Become A 24-Point Master

算24点。傻逼了，7个7竟然没算出来。。贴纬哥的代码

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<string>
  5 #include<cstring>
  6 #include<algorithm>
  7 #include<set>
  8 using namespace std;
  9 
 10 
 11 #define inf 0x3f3f3f3f
 12 #define eps 1e-9
 13 #define FOR(i,s,t) for(int i = s; i < t; ++i )
 14 #define REP(i,s,t) for( int i = s; i <= t; ++i )
 15 #define pii pair<int,int>
 16 #define MP make_pair
 17 #define lson id << 1 , l , mid
 18 #define rson id << 1 | 1 , mid + 1 , r
 19 #define LL long long
 20 #define ULL unsigned long long
 21 #define N ( 500000 + 10  )
 22 #define M ( 1000000 + 10)
 23 #define mod 1000000007
 24 
 25 
 26 void out ( int n )
 27 {
 28 
 29     if( n <= 3 ) {
 30         puts("-1");
 31         return ;
 32     }
 33     if( n == 4 ) {
 34         string s = "1 * 2\n3 + 4\n5 + 6";
 35         cout<<s<<endl;
 36     }
 37     if( n == 5 ) {
 38         puts("1 / 2");
 39         puts("6 / 3");
 40         puts("4 - 7");
 41         puts("5 * 8");
 42     }
 43     if( n == 6 ) {
 44         puts("1 - 2");
 45         printf("%d + %d\n%d + %d\n", 3, 4, 5, 6 );
 46         for( int i = 7; i < 11; i+=2 )
 47             printf("%d + %d\n", i, i + 1 );
 48     }
 49     if( n == 7 ) {
 50  puts("1 * 2"); // 8 = 49
 51         puts("3 / 4"); // 9 = 1
 52         puts("8 - 9"); // 10 = 48
 53         puts("5 + 6");
 54         puts("10 / 11");
 55         puts("12 * 7");    }
 56     if( n == 8 ) {
 57         puts("1 + 2");
 58         puts("9 + 3"); // 10 = 24
 59         puts("4 - 5"); // 11 = 0
 60         int i = 6, j = 11;
 61         for( ; i <= 8; ++i )
 62             printf("%d * %d\n", i, j ++ );
 63         printf("%d + %d\n",j, 10);
 64     }
 65     if( n == 9 ) {
 66         for( int i = 1; i <= 6; i+=2 )
 67             printf("%d / %d\n", i, i + 1 );
 68         puts("7 + 8");
 69         printf("%d + %d\n",13, 9 ); //a[14] = 27
 70         int i = 14, j = 10;
 71         while( j <= 12 ) printf("%d - %d\n", i++, j++ );
 72     }
 73     if( n == 10 )
 74     {
 75         int k = 10;
 76         for( int i = 1; i <= 8; i += 2 )
 77             printf("%d / %d\n", i, i + 1 ), ++k;
 78         printf("%d + %d\n", 9, 10 );++k;
 79         int j = 11;
 80         while( k < 19 )
 81             printf("%d + %d\n", k++, j++);
 82     }
 83     if( n == 11 ) {
 84         puts("1 + 2"); // 12
 85         puts("12 / 3"); // 13
 86         puts("4 + 5"); // 14
 87         puts("6 - 7"); // 15
 88         int j = 15;
 89         for( int i = 8; i <= 11; ++i )
 90             printf("%d * %d\n", j++, i );
 91         for( int i = 13; i <= 14; ++i )
 92             printf("%d + %d\n", i, j ++);
 93     }
 94     if( n == 12 ) {
 95         puts("1 + 2"); // 13
 96         puts("3 - 4"); // 14
 97         int j = 14;
 98         for( int i = 5; i <= 12; ++i )
 99             printf("%d * %d\n", i, j ++ );
100         printf("%d + %d\n", j, 13 );
101     }
102     if(n == 13 ) {
103         puts("1 + 2");
104         puts("3 + 4" );
105         puts("15 / 5");//16
106         puts("14 - 16"); // 17 = 24;
107         puts("6 - 7"); // 18
108         int j = 18;
109         for( int i = 8 ; i <= 13; i ++ )
110             printf("%d * %d\n", j++, i );
111         printf("%d + %d\n", j , 17 );
112     }
113     if( n ==15 ) {
114         puts("1 + 2"); //16 30
115         puts("16 / 3"); // 17 2
116         puts("4 + 5"); // 18 2n
117         puts("18 / 6"); // 19 2
118         puts("7 / 8"); // 20 1;
119         puts("20 + 19"); // 21 3
120         puts("17 * 21"); // 22 6;
121         puts("9 + 10"); // 23 30;
122         puts("23 - 22"); // 24 = 24
123         puts("11 - 12"); // 25 = 0;
124         int j = 25;
125         for( int i = 13; i <= 15; ++i )
126             printf("%d * %d\n", i, j++ );
127         printf("%d + %d\n", j, 24 );
128     }
129 
130 
131 }
132 
133 char s[100];
134 int A[1000];
135 bool vis[1000];
136 void check( )
137 {
138     int n;
139     while( cin>>n ) {
140         getchar();
141         for( int i = 1; i <= n; ++i )
142             A[i] = n;
143         memset( vis, 0, sizeof( vis ));
144         bool e = 1;
145         int k = n + 1;
146         for( int z = 1; z <=  n - 1; ++z ) {
147             gets(s);
148             int len = strlen( s );
149             int a = 0, b =  0;
150             int i;
151             for( i = 0; i < len; ++i ) {
152                 if( '0'<= s[i] && s[i] <= '9')
153                     a = a * 10 + s[i] - '0';
154                 else break;
155             }
156             int j = i + 1;
157             for( i = i + 3; i < len; ++i )
158                 b = b * 10 + s[i] - '0';
159 
160             if( vis[a] ) e = 0;
161             if( vis[b] ) e = 0;
162             if( a >= k || b >= k ) e = 0;
163             vis[a] = vis[b] = 1;
164             if( s[j] == '+') A[k++] = A[a] + A[b];
165             if( s[j] == '-') A[k++] = A[a] - A[b];
166             if( s[j] == '*' ) A[k++] = A[a] * A[b];
167             if( s[j] == '/') A[k++] = A[a] / A[b];
168         }
169         for( int i = 1; i <= 2 * n -1; ++i )
170             printf("%d ", A[i] );
171         puts("");
172 
173         if( e && A[2*n-1] == 24 )
174             puts("YES");
175             else {
176                 printf("%d NO\n", n );
177             }
178     }
179 }
180 int main ( )
181 {
182 
183     int n;
184 
185 
186     while(~scanf("%d", &n ) ) {
187         if( n >= 14 && n != 15 ) {
188             for( int i = 1; i <= 7; i += 2 )
189                 printf("%d + %d\n", i, i + 1 );
190             for( int i = 1; i <= 4; ++i )
191                 printf("%d / %d\n", n+i, i + 8 );
192             printf("%d / %d\n", 13, 14 );// n + 9 == 1
193 
194             printf("%d + %d\n", n + 5, n + 6 ); // a[n+10] = 4
195             printf("%d + %d\n", n + 7, n + 9 ); // a[n+11] = 3
196             printf("%d * %d\n", n + 11, n + 8 ); // a[n+12] = 6;
197 
198                 printf("%d * %d\n", n+10, n + 12 ); // a[n+13] = 24;
199 
200             if( n == 14 ) continue;
201             printf("%d - %d\n", 15, 16 ); //a[n+14] = 0;
202             int  j = n + 14;
203             for( int i = 17; i <= n; ++i )
204                 printf("%d * %d\n", i, j++ );
205             printf("%d + %d\n", j, n + 13 );
206 
207         }
208         else out( n );
209     }
210 
211 }
View Code

 

posted @ 2015-07-30 02:05  L__J  阅读(209)  评论(0)  编辑  收藏  举报