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<>J 翻转硬币
不会写
<>I 像素
枚举第i行
二进制枚举状态
然后check(i)是否合法,如果合法就dfs(i+1)
check是核心
判断第上一行是否==A[i][j]
判断第i行是否小于等于A且,c+3>=A
判断下一行是否小于等于A且,c+6>=A
<>H 异或和求和
按位做就好了
比如
5
1 2 3 4 5
bit=0 数组变成 10101
bit=1 数组变成 01100
单独考虑bit=0,模2意义下求前缀和变成11001
开一个数组c[2]
这一位的贡献就变成了sum+=c[1-b[i]]
然后记录一下c[b[i]]++
ans+=(1<<bit)*sum
<>G
克鲁斯卡尔重构一个最大树,对于每个询问求lca
刚开始没想到,我还想二分答案判断是否能用权值大于mid的边构造u到v的连通性
<>F
不会,我打的暴力
盲猜一手折半搜索可以处理
<>E
树上启发式合并
维护一个ma[u]和sz[u]
ma表示最大颜色个数
sz表示size
mp[u].size()表示颜色段
判断(sz%mp.size()==0 && sz/mp.size()==ma) ans++
<>D
区间dp预处理一下
for(len=2~n) for(l=1~) int r=l+len-1; if(s[l]>s[r])f[l][r]=1; else if(s[l]==s[r
])f[l][r]=f[l+1][r-1] for(i=2~n) for(j) ans+=f[j][i]; cout<<ans