标签：
1. All X
1.1 基本思路
k和c的范围都不大，因此可以考虑迭代找循环节，然后求余数，判定是否相等。这题还是挺简单的。
1.2 代码
1 /* 5690 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 44 typedef long long LL; 45 const int maxn = 10005; 46 int x; 47 LL m; 48 int c, k; 49 int visit[maxn]; 50 51 int main() { 52 ios::sync_with_stdio(false); 53 #ifndef ONLINE_JUDGE 54 freopen("data.in", "r", stdin); 55 freopen("data.out", "w", stdout); 56 #endif 57 58 int t; 59 int tmp, ntmp; 60 LL i, cycle; 61 62 scanf("%d", &t); 63 rep(tt, 1, t+1) { 64 scanf("%d%I64d%d%d", &x,&m,&k,&c); 65 printf("Case #%d:\n", tt); 66 memset(visit, -1, sizeof(visit)); 67 ntmp = x % k; 68 cycle = -1; 69 for (i=1; i<=m; ++i) { 70 if (visit[tmp=ntmp] != -1) { 71 cycle = i - visit[tmp]; 72 break; 73 } 74 visit[tmp] = i; 75 ntmp = (10*tmp + x) % k; 76 } 77 78 if (cycle == -1) { 79 puts(tmp==c ? "Yes":"No"); 80 } else { 81 LL n = ((m - visit[tmp]) % cycle + cycle) % cycle; 82 for (i=0; i<n; ++i) 83 tmp = (10*tmp + x) % k; 84 puts(tmp==c ? "Yes":"No"); 85 } 86 } 87 88 #ifndef ONLINE_JUDGE 89 printf("time = %ldms.\n", clock()); 90 #endif 91 92 return 0; 93 }
2. Sitting in Line
2.1 基本思路
N的范围很小，刚开始以为这是一个数学题，后来发现状态DP可解。
$dp[st][n]$st表示当前状态（即当前已经加入数字），n表示当前排列的最末端数字。显然有状态转移方程
\[
dp[nst][k] = \max(dp[nst][k], dp[st|(1<<k)][j]+a_j \times a_k), \\
\qquad \text{其中 } st\&(1<<k) == 0\text{ 并且 }pos_k==-1 || pos_k==getBits(st)
\]
2.2 代码
1 /* 5691 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 44 typedef long long LL; 45 LL INF = 0x3f3f3f3f3f3f3f3f; 46 LL NEG_INF = 0xc0c0c0c0c0c0c0c0; 47 const int maxn = 16; 48 LL dp[1<<maxn][maxn]; 49 int a[maxn], p[maxn]; 50 int Bits[1<<maxn]; 51 int n; 52 53 int getBits(int x) { 54 int ret = 0; 55 56 while (x) { 57 ret += x & 1; 58 x >>= 1; 59 } 60 61 return ret; 62 } 63 64 void init() { 65 const int mst = 1<<16; 66 rep(i, 0, mst) Bits[i] = getBits(i); 67 } 68 69 void solve() { 70 memset(dp, 0xC0, sizeof(dp)); 71 rep(i, 0, n) { 72 if (p[i]==-1 || p[i]==0) 73 dp[1<<i][i] = 0; 74 } 75 76 const int mst = 1<<n; 77 #ifndef ONLINE_JUDGE 78 printf("NEG_INF = %I64d\n", NEG_INF); 79 #endif 80 rep(i, 0, mst) { 81 const int idx = Bits[i]; 82 rep(j, 0, n) { 83 if (dp[i][j] == NEG_INF) continue; 84 rep(k, 0, n) { 85 if (i & (1<<k)) continue; 86 if (p[k]==-1 || p[k]==idx) { 87 int nst = i | 1<<k; 88 dp[nst][k] = max(dp[nst][k], dp[i][j]+a[j]*a[k]); 89 } 90 } 91 } 92 } 93 94 LL ans = NEG_INF; 95 rep(j, 0, n) 96 ans = max(ans, dp[mst-1][j]); 97 printf("%I64d\n", ans); 98 } 99 100 int main() { 101 ios::sync_with_stdio(false); 102 #ifndef ONLINE_JUDGE 103 freopen("data.in", "r", stdin); 104 freopen("data.out", "w", stdout); 105 #endif 106 107 int t; 108 109 init(); 110 scanf("%d", &t); 111 rep(tt, 1, t+1) { 112 scanf("%d", &n); 113 rep(i, 0, n) 114 scanf("%d%d", &a[i],&p[i]); 115 printf("Case #%d:\n", tt); 116 solve(); 117 } 118 119 #ifndef ONLINE_JUDGE 120 printf("time = %ldms.\n", clock()); 121 #endif 122 123 return 0; 124 }
3. Snacks
3.1 基本思路
这个题刚开始看以为是树链剖分，巨难无比，赛后发现就是个树形结构转线性结构，然后使用线段树维护最大值就好了，使用下lazy标记直接A了。
3.2 代码
1 /* 5692 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 #pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 44 struct edge_t { 45 int v, nxt; 46 }; 47 48 typedef long long LL; 49 const int maxn = 1e5+5; 50 const int maxv = maxn; 51 const int maxe = maxn * 2; 52 int head[maxv], l, cnt; 53 edge_t E[maxe]; 54 LL mx[maxn<<2], delta[maxn<<2]; 55 int L[maxn], R[maxn], a[maxn]; 56 int LLL, RR, w; 57 int n, m; 58 59 void init() { 60 memset(head, -1, sizeof(head)); 61 l = cnt = 0; 62 memset(mx, 0, sizeof(mx)); 63 memset(delta, 0, sizeof(delta)); 64 } 65 66 void addEdge(int u, int v) { 67 E[l].v = v; 68 E[l].nxt = head[u]; 69 head[u] = l++; 70 71 E[l].v = u; 72 E[l].nxt = head[v]; 73 head[v] = l++; 74 } 75 76 void dfs(int u, int fa) { 77 L[u] = ++cnt; 78 for (int k=head[u]; k!=-1; k=E[k].nxt) { 79 const int& v = E[k].v; 80 if (v == fa) continue; 81 dfs(v, u); 82 } 83 R[u] = cnt; 84 } 85 86 inline void PushDown(int rt) { 87 if (delta[rt]) { 88 int lb = rt<<1, rb = lb | 1; 89 delta[lb] += delta[rt]; 90 delta[rb] += delta[rt]; 91 mx[lb] += delta[rt]; 92 mx[rb] += delta[rt]; 93 delta[rt] = 0; 94 } 95 } 96 97 inline void PushUp(int rt) { 98 mx[rt] = max(mx[rt<<1], mx[rt<<1|1]); 99 } 100 101 void Update(int l, int r, int rt) { 102 if (LLL<=l && RR>=r) { 103 mx[rt] += w; 104 delta[rt] += w; 105 return ; 106 } 107 108 PushDown(rt); 109 int mid = (l + r) >> 1; 110 111 if (RR <= mid) { 112 Update(lson); 113 } else if (LLL > mid) { 114 Update(rson); 115 } else { 116 Update(lson); 117 Update(rson); 118 } 119 120 PushUp(rt); 121 } 122 123 LL Query(int l, int r, int rt) { 124 if (LLL<=l && RR>=r) { 125 return mx[rt]; 126 } 127 128 PushDown(rt); 129 int mid = (l + r) >> 1; 130 131 if (RR <= mid) 132 return Query(lson); 133 else if (LLL > mid) 134 return Query(rson); 135 else 136 return max(Query(lson), Query(rson)); 137 } 138 139 void solve() { 140 dfs(0, -1); 141 142 rep(i, 0, n) { 143 scanf("%d", &a[i]); 144 LLL = L[i]; 145 RR = R[i]; 146 w = a[i]; 147 Update(1, n, 1); 148 } 149 150 int op, x, y; 151 LL ans; 152 153 rep(i, 0, m) { 154 scanf("%d%d", &op, &x); 155 LLL = L[x]; 156 RR = R[x]; 157 if (op) { 158 ans = Query(1, n, 1); 159 printf("%I64d\n", ans); 160 } else { 161 scanf("%d", &y); 162 y -= a[x]; 163 w = y; 164 Update(1, n, 1); 165 a[x] += y; 166 } 167 } 168 } 169 170 int main() { 171 ios::sync_with_stdio(false); 172 #ifndef ONLINE_JUDGE 173 freopen("data.in", "r", stdin); 174 freopen("data.out", "w", stdout); 175 #endif 176 177 int t; 178 int u, v; 179 180 scanf("%d", &t); 181 rep(tt, 1, t+1) { 182 scanf("%d%d", &n,&m); 183 init(); 184 rep(i, 1, n) { 185 scanf("%d%d", &u,&v); 186 addEdge(u, v); 187 } 188 printf("Case #%d:\n", tt); 189 solve(); 190 } 191 192 #ifndef ONLINE_JUDGE 193 printf("time = %ldms.\n", clock()); 194 #endif 195 196 return 0; 197 }
4. D Game
4.1 基本思路
$n,m \in [1,300]$这数据范围很小，基本思路是DP.
不妨令$dp[i][j]$表示从第$i$个数字到第$j$个数字能否通过全部被删掉，$mx[i]$表示前$i$个数字中可以最多可以删掉的数字。显然有动态转移方程
\[
mx[i] = \max(mx[i], mx[j-1]+i-j+1), \text{ if } dp[j][i]=True
\]
因此，紧急考虑$dp[j][i]$的情况即可：
(1) $dp[j][k] \& dp[k+1][i], k \in [j+1, i)$
(2) $dp[j+1][i-1], (a_i-a_j) \in D$
(3) $dp[j+1][k-1] \& dp[k+1][i-1], (a_i-a_j)/2 \in D, a_k*2 = a_i+a_j, k \in [j+1, i)$
4.2 代码
1 /* 5693 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 44 const int maxn = 305; 45 bool dp[maxn][maxn]; 46 int mx[maxn]; 47 int a[maxn]; 48 int n, m; 49 sti st; 50 51 void solve() { 52 memset(dp, 0, sizeof(dp)); 53 54 rep(i, 1, n+1) { 55 dp[i][i] = false; 56 dp[i][i-1] = true; 57 per(j, 1, i) { 58 rep(k, j+1, i) 59 dp[j][i] |= dp[j][k] & dp[k+1][i]; 60 if (st.count(a[i]-a[j])) { 61 dp[j][i] |= dp[j+1][i-1]; 62 } 63 if ((a[i]-a[j])%2==0 && st.count((a[i]-a[j])>>1)) { 64 rep(k, j+1, i) { 65 if (a[k]+a[k] == a[j]+a[i]) { 66 dp[j][i] |= dp[j+1][k-1] & dp[k+1][i-1]; 67 } 68 } 69 } 70 } 71 } 72 73 mx[0] = 0; 74 rep(i, 1, n+1) { 75 mx[i] = mx[i-1]; 76 rep(j, 1, i) { 77 if (dp[j][i]) 78 mx[i] = max(mx[i], mx[j-1]+i-j+1); 79 } 80 } 81 82 printf("%d\n", mx[n]); 83 } 84 85 int main() { 86 ios::sync_with_stdio(false); 87 #ifndef ONLINE_JUDGE 88 freopen("data.in", "r", stdin); 89 freopen("data.out", "w", stdout); 90 #endif 91 92 int t; 93 int x; 94 95 scanf("%d", &t); 96 while (t--) { 97 scanf("%d%d", &n,&m); 98 rep(i, 1, n+1) scanf("%d", a+i); 99 st.clr(); 100 rep(i, 0, m) { 101 scanf("%d", &x); 102 st.insert(x); 103 } 104 solve(); 105 } 106 107 #ifndef ONLINE_JUDGE 108 printf("time = %ldms.\n", clock()); 109 #endif 110 111 return 0; 112 }
5. BD String
5.1 基本思路
不妨令$f(n)$表示$S(n)$字符串的长度，显然很容易得到$f(n) = 2^n-1$。因此，最终串的长度为$f(2^{1000}) = 2^{1001}-1$，而$L,R \in [1,10^{18}]$。
显然原问题等价于计算$S(60)$的$[L,R]$范围内的B的个数。
算法很简单，就是个递归。可以考虑$S(k)$的左半部分和右半部分，递归时，加上剪枝$R-L+1==f(dep)$即可，此时B的数量为$f(dep-1)$,D的数量为$f(dep-1)-1$。
5.2 代码
1 /* 5694 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 #pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 44 typedef long long LL; 45 LL Base[62]; 46 LL L, R; 47 48 void init() { 49 rep(i, 0, 62) 50 Base[i] = 1LL << i; 51 } 52 53 LL dfsD(LL, LL, int); 54 LL dfsB(LL, LL, int); 55 56 LL dfsD(LL L, LL R, int dep) { 57 if (L > R) return 0; 58 if (dep == 0) return 0; 59 if (dep == 1) return 0; 60 61 if (R-L+1 == Base[dep]-1) { 62 return Base[dep-1]-1; 63 } 64 LL mid = Base[dep-1]; 65 if (R < mid) { 66 return dfsD(L, R, dep-1); 67 } else if (L > mid) { 68 LL nxtL = mid - 1 - (R - mid) + 1; 69 LL nxtR = mid - 1 - (L - mid) + 1; 70 return dfsB(nxtL, nxtR, dep-1); 71 72 } else { 73 LL lret = dfsD(L, mid-1, dep-1); 74 LL rret = 0; 75 76 if (R > mid) { 77 LL nxtL = mid-1 - (R-mid) + 1; 78 LL nxtR = mid-1; 79 rret = dfsB(nxtL, nxtR, dep-1); 80 } 81 82 return lret + rret; 83 } 84 } 85 86 LL dfsB(LL L, LL R, int dep) { 87 if (L > R) return 0; 88 if (dep == 0) return 0; 89 if (dep == 1) return 1; 90 91 if (R-L+1 == Base[dep]-1) { 92 return Base[dep-1]; 93 } 94 LL mid = Base[dep-1]; 95 if (R < mid) { 96 return dfsB(L, R, dep-1); 97 98 } else if (L > mid) { 99 LL nxtL = mid - 1 - (R - mid) + 1; 100 LL nxtR = mid - 1 - (L - mid) + 1; 101 return dfsD(nxtL, nxtR, dep-1); 102 103 } else { 104 LL lret = dfsB(L, mid-1, dep-1); 105 LL rret = 0; 106 107 if (R > mid) { 108 LL nxtL = mid-1 - (R-mid) + 1; 109 LL nxtR = mid-1; 110 rret = dfsD(nxtL, nxtR, dep-1); 111 } 112 113 return lret + 1 + rret; 114 } 115 } 116 117 void solve() { 118 LL ans = dfsB(L, R, 60); 119 120 printf("%I64d\n", ans); 121 } 122 123 int main() { 124 ios::sync_with_stdio(false); 125 #ifndef ONLINE_JUDGE 126 freopen("data.in", "r", stdin); 127 freopen("data.out", "w", stdout); 128 #endif 129 130 int t; 131 132 init(); 133 scanf("%d", &t); 134 while (t--) { 135 scanf("%I64d%I64d", &L, &R); 136 solve(); 137 } 138 139 #ifndef ONLINE_JUDGE 140 printf("time = %ldms.\n", clock()); 141 #endif 142 143 return 0; 144 }
6. Gym Class
6.1 基本思路
水题，根据依赖性建图，然后拓扑排序就好了。注意拓扑的时候使用优先队列可求最值。
6.2 代码
1 /* */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <bitset> 12 #include <algorithm> 13 #include <cstdio> 14 #include <cmath> 15 #include <ctime> 16 #include <cstring> 17 #include <climits> 18 #include <cctype> 19 #include <cassert> 20 #include <functional> 21 #include <iterator> 22 #include <iomanip> 23 using namespace std; 24 #pragma comment(linker,"/STACK:102400000,1024000") 25 26 #define sti set<int> 27 #define stpii set<pair<int, int> > 28 #define mpii map<int,int> 29 #define vi vector<int> 30 #define pii pair<int,int> 31 #define vpii vector<pair<int,int> > 32 #define rep(i, a, n) for (int i=a;i<n;++i) 33 #define per(i, a, n) for (int i=n-1;i>=a;--i) 34 #define clr clear 35 #define pb push_back 36 #define mp make_pair 37 #define fir first 38 #define sec second 39 #define all(x) (x).begin(),(x).end() 40 #define SZ(x) ((int)(x).size()) 41 #define lson l, mid, rt<<1 42 #define rson mid+1, r, rt<<1|1 43 44 struct edge_t { 45 int v, nxt; 46 }; 47 48 typedef long long LL; 49 const int maxn = 1e5+5; 50 const int maxv = maxn; 51 const int maxe = maxn; 52 int head[maxv], l; 53 edge_t E[maxe]; 54 int deg[maxv]; 55 bool visit[maxv]; 56 int n, m; 57 58 void init() { 59 memset(head, -1, sizeof(head)); 60 memset(deg, 0, sizeof(deg)); 61 memset(visit, false, sizeof(visit)); 62 l = 0; 63 } 64 65 inline void addEdge(int u, int v) { 66 ++deg[v]; 67 E[l].v = v; 68 E[l].nxt = head[u]; 69 head[u] = l++; 70 } 71 72 void solve() { 73 priority_queue<int> Q; 74 LL ans = 0; 75 int u, v, k, mn = INT_MAX; 76 77 rep(i, 1, n+1) { 78 if (deg[i] == 0) { 79 Q.push(i); 80 visit[i] = true; 81 } 82 } 83 84 while (!Q.empty()) { 85 u = Q.top(); 86 Q.pop(); 87 mn = min(u, mn); 88 ans += mn; 89 for (k=head[u]; k!=-1; k=E[k].nxt) { 90 v = E[k].v; 91 if (!visit[v] && --deg[v]==0) { 92 Q.push(v); 93 visit[v] = true; 94 } 95 } 96 } 97 98 printf("%I64d\n", ans); 99 } 100 101 int main() { 102 ios::sync_with_stdio(false); 103 #ifndef ONLINE_JUDGE 104 freopen("data.in", "r", stdin); 105 freopen("data.out", "w", stdout); 106 #endif 107 108 int t; 109 int u, v; 110 111 scanf("%d", &t); 112 while (t--) { 113 scanf("%d%d", &n, &m); 114 init(); 115 rep(i, 0, m) { 116 scanf("%d%d", &u,&v); 117 addEdge(u, v); 118 } 119 solve(); 120 } 121 122 #ifndef ONLINE_JUDGE 123 printf("time = %ldms.\n", clock()); 124 #endif 125 126 return 0; 127 }
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原文地址：http://www.cnblogs.com/bombe1013/p/5515721.html
