Codeforces Round #287 (Div. 2) ABCDE
A题。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define inf 0x3f3f3f3f 13 #define lson l, m, rt << 1 14 #define rson m+1, r, rt << 1 | 1 15 #define mnx 500 16 17 struct node{ 18 int v, id; 19 bool operator < ( const node &b ) const { 20 return v < b.v; 21 } 22 }a[mnx]; 23 int b[mnx]; 24 int main(){ 25 int n, k; 26 scanf( "%d %d", &n, &k ); 27 for( int i = 1; i <= n; ++i ){ 28 scanf( "%d", &a[i].v ); 29 a[i].id = i; 30 } 31 sort( a + 1, a + 1 + n ); 32 int i = 1, ans = 0, sum = 0; 33 while( sum + a[i].v <= k && i <= n ){ 34 b[ans++] = a[i].id, sum += a[i++].v; 35 } 36 printf( "%d\n", ans ); 37 for( i = 0; i < ans; ++i ) 38 printf( "%d ", b[i] ); 39 return 0; 40 }
B题。。注意有可能 爆int
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define inf 0x3f3f3f3f 13 #define lson l, m, rt << 1 14 #define rson m+1, r, rt << 1 | 1 15 #define mnx 500 16 17 int main(){ 18 double r, x, y, xx, yy; 19 scanf( "%lf%lf%lf%lf%lf", &r, &x, &y, &xx, &yy ); 20 double dis = sqrt( ( xx - x ) * ( xx - x ) + ( yy - y ) * ( yy - y ) ); 21 int ans = ceil( dis / 2 / r ); 22 cout << ans << endl; 23 }
C题。。把它当做一棵满二叉树来看，根节点是1，然后第二层是2,3， 第三层是4,5,6,7。。先计算出第h层第n个节点在整个满二叉树是第cur个，然后一直向上得到父亲节点，一直到根节点。。计算的话，因为是LRLRLR这样走的，如果a[i+1] %2 == a[i]%2的话，就要加上某一边的子树的所有节点b[h+1-i]，不同的话就直接ans++就好。。具体自己画图加结合代码理解一下
1 include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 #include <vector> 6 7 using namespace std; 8 9 #define LL long long 10 #define eps 1e-8 11 #define inf 0x3f3f3f3f 12 #define lson l, m, rt << 1 13 #define rson m+1, r, rt << 1 | 1 14 #define mnx 500 15 16 LL a[mnx], b[mnx]; 17 int main(){ 18 LL h, n; 19 cin >> h >> n; 20 LL L = 1; 21 for( int i = 0; i < h; ++i ) 22 L <<= 1; 23 b[0] = 1; 24 for( int i = 1; i <= h; ++i ) 25 b[i] = b[i-1] * 2; 26 LL cur = L + n - 1; 27 int m = h + 1; 28 while( cur ){ 29 a[m--] = cur; 30 cur >>= 1; 31 } 32 LL ans = 0; 33 for( int i = 1; i < h+1; ++i ){ 34 if( a[i+1] % 2 == a[i] % 2 ) 35 ans += b[h+1-i]; 36 else ans++; 37 } 38 cout << ans << endl; 39
D题。。不会，数位dp，复制别人的题解
给定n，k，m
求有多少个恰好为n位的整数，且这个整数的某个后缀能整除k，答案模m
思路：
因为是求后缀，所以从低位往高位构造。
dp[i][j][0] 表示长度为i位的整数，模k结果为j，且不存在某个后缀能整除k的个数
dp[i][j][1] 表示长度为i位的整数，模k结果为j，且存在某个后缀能整除k的个数
然后转移就在某个状态前加一个数字，判断一下加上后能否被k整除来决定加上数字后到达的状态
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 #include <map> 8 #include <queue> 9 10 using namespace std; 11 12 #define LL long long 13 #define eps 1e-8 14 #define inf 0x3f3f3f3f 15 #define lson l, m, rt << 1 16 #define rson m+1, r, rt << 1 | 1 17 #define mnx 1050 18 19 LL dp[mnx][120][2], len[mnx], mod; 20 int n, k; 21 void add( LL &x, LL y ){ 22 x = ( x + y ) % mod; 23 } 24 int main(){ 25 cin >> n >> k >> mod; 26 dp[0][0][0] = 1, len[0] = 1; 27 for( int i = 1; i < mnx; ++i ) 28 len[i] = len[i-1] * 10 % k; 29 for( int i = 0; i < n; ++i ){ 30 for( int j = 0; j < k; ++j ){ 31 for( int v = 0; v < 10; ++v ){ 32 if( i == n-1 && !v ) continue; 33 LL val = ( v * len[i] + j ) % k; 34 if( !val && v ) 35 add( dp[i+1][0][1], dp[i][j][0] ); 36 else 37 add( dp[i+1][val][0], dp[i][j][0] ); 38 add( dp[i+1][val][1], dp[i][j][1] ); 39 } 40 } 41 } 42 LL ans = 0; 43 for( int i = 0; i < k; ++i ) 44 ans = ( ans + dp[n][i][1] ) % mod; 45 cout << ans << endl; 46 return 0; 47 }
Ｅ题。。最短路，感觉挺水的，把最短路上所有为0的边改为1，把不是最短路的边所有为1的变为0。。
表示自己的代码写的好挫。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <vector> 7 #include <map> 8 #include <queue> 9 10 using namespace std; 11 12 #define LL long long 13 #define eps 1e-8 14 #define inf 0x3f3f3f3f 15 #define lson l, m, rt << 1 16 #define rson m+1, r, rt << 1 | 1 17 #define mnx 200500 18 19 int fst[mnx], nxt[mnx], vv[mnx], work[mnx], e, dis[mnx], sum[mnx], n, fa[mnx]; 20 bool vis[mnx]; 21 map< pair<int,int>, int > mp; 22 void add( int u, int v, int c ){ 23 vv[e] = v, nxt[e] = fst[u], work[e] = c, fst[u] = e++; 24 } 25 struct edge{ 26 int v, u, d; 27 edge(){} 28 edge( int u, int v, int d ) : u(u), v(v), d(d) {} 29 bool operator < ( const edge &b ) const { 30 return d > b.d; 31 } 32 }ee[mnx]; 33 void dij(){ 34 memset( fa, -1, sizeof fa ); 35 fa[1] = 1; 36 memset( dis, 0x3f, sizeof dis ); 37 priority_queue<edge> q; 38 dis[1] = 0; 39 edge p; 40 p.u = 1, p.d = 0; 41 q.push( p ); 42 while( !q.empty() ){ 43 p = q.top(); q.pop(); 44 int u = p.u; 45 if( vis[u] ) continue; 46 vis[u] = 1; 47 for( int i = fst[u]; i != -1; i = nxt[i] ){ 48 int v = vv[i], w = work[i]; 49 if( dis[v] > dis[u] + 1 || ( dis[v] == dis[u] + 1 && sum[v] < sum[u] + w ) ){ 50 dis[v] = dis[u] + 1; 51 sum[v] = sum[u] + w; 52 edge pp; 53 pp.u = v, pp.d = dis[v]; 54 q.push( pp ); 55 fa[v] = u; 56 } 57 } 58 } 59 //cout << dis[n] << endl; 60 } 61 void dfs( int u ){ 62 vis[u] = 1; 63 if( fa[u] != u ){ 64 dfs( fa[u] ); 65 } 66 } 67 int main(){ 68 int m, all = 0; 69 scanf( "%d%d",&n, &m ); 70 memset( fst, -1, sizeof fst ); 71 for( int i = 0; i < m; ++i ){ 72 int u, v, c; 73 scanf( "%d%d%d", &u, &v, &c ); 74 add( u, v, c ); 75 add( v, u, c ); 76 ee[all++] = edge( u, v, c ); 77 ee[all++] = edge( v, u, c ); 78 } 79 dij(); 80 memset( vis, 0, sizeof vis ); 81 dfs( n ); 82 int ans = 0; 83 for( int i = 1; i <= n; ++i ){ 84 int u = i; 85 if( !vis[i] ){ 86 for( int j = fst[u]; j != -1; j = nxt[j] ){ 87 int v = vv[j]; 88 if( mp.find( make_pair( u, v ) ) == mp.end() && mp.find( make_pair( v, u ) ) == mp.end() ){ 89 if( work[j] ) ans++; 90 mp[make_pair(u,v)]++; 91 } 92 } 93 } 94 else{ 95 for( int j = fst[u]; j != -1; j = nxt[j] ){ 96 int v = vv[j]; 97 if( v == fa[u] ){ 98 if( !work[j] ) ans++; 99 mp[make_pair(u,v)]++; 100 } 101 } 102 } 103 } 104 for( int i = 1; i <= n; ++i ){ 105 int u = i; 106 if( vis[i] ){ 107 for( int j = fst[u]; j != -1; j = nxt[j] ){ 108 int v = vv[j]; 109 if( mp.find( make_pair(u,v) ) == mp.end() && mp.find( make_pair(v,u) ) == mp.end() ){ 110 if( work[j] ) ans++; 111 mp[make_pair(u,v)]++; 112 } 113 } 114 } 115 } 116 cout << ans << endl; 117 mp.clear(); 118 for( int i = 1; i <= n; ++i ){ 119 int u = i; 120 if( !vis[i] ){ 121 for( int j = fst[u]; j != -1; j = nxt[j] ){ 122 int v = vv[j]; 123 if( mp.find( make_pair( u, v ) ) == mp.end() && mp.find( make_pair( v, u ) ) == mp.end() ){ 124 if( work[j] ) 125 printf( "%d %d %d\n", min(u,v),max(u,v), 0 ); 126 mp[make_pair(u,v)]++; 127 } 128 } 129 } 130 else{ 131 for( int j = fst[u]; j != -1; j = nxt[j] ){ 132 int v = vv[j]; 133 if( v == fa[u] ){ 134 if( !work[j] ) 135 printf( "%d %d %d\n", min(u,v),max(u,v), 1 ); 136 mp[make_pair(u,v)]++; 137 } 138 } 139 } 140 } 141 for( int i = 1; i <= n; ++i ){ 142 int u = i; 143 if( vis[i] ){ 144 for( int j = fst[u]; j != -1; j = nxt[j] ){ 145 int v = vv[j]; 146 if( mp.find( make_pair(u,v) ) == mp.end() && mp.find( make_pair(v,u) ) == mp.end() ){ 147 if( work[j] ) 148 printf( "%d %d %d\n", min(u,v),max(u,v), 0 ); 149 mp[make_pair(u,v)]++; 150 } 151 } 152 } 153 } 154 return 0; 155 }
