【gym102394B】Binary Numbers（DP）

Meaning of the title: the From  https://blog.csdn.net/m0_37809890/article/details/102886956

 

 Ideas:

 

Transfer can be found in the upper right corner is a range of prefixes and

std 1 times as long as the open space, personal habits open 2 times rolling array initialization sequence or seeking prefix and tremor will be doubled under gg sometimes thought clear premise, anyway, only a constant multiple spatial and temporal gap

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef long double ld;
  7 typedef pair<int,int> PII;
  8 typedef pair<ll,ll> Pll;
  9 typedef vector<int> VI;
 10 typedef vector<PII> VII;
 11 typedef pair<ll,ll>P;
 12 #define N  1<<20
 13 #define M  1000000
 14 #define INF 1e9
 15 #define fi first
 16 #define se second
 17 #define MP make_pair
 18 #define pb push_back
 19 #define pi acos(-1)
 20 #define mem(a,b) memset(a,b,sizeof(a))
 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 23 #define lowbit(x) x&(-x)
 24 #define Rand (rand()*(1<<16)+rand())
 25 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 26 #define ls p<<1
 27 #define rs p<<1|1
 28 #define fors(i) for(auto i:e[x]) if(i!=p)
 29 
 30 const int MOD=1e8+7,inv2=(MOD+1)/2;
 31       double eps=1e-6;
 32       int dx[4]={-1,1,0,0};
 33       int dy[4]={0,0,-1,1};
 34 
 35 ll s[2][20][20],dp[2][20][20];
 36 int l[N],r[N],m,n;
 37 
 38 int read()
 39 {
 40    int v=0,f=1;
 41    char c=getchar();
 42    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 43    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 44    return v*f;
 45 }
 46 
 47 ll readll()
 48 {
 49    ll v=0,f=1;
 50    char c=getchar();
 51    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 52    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 53    return v*f;
 54 }
 55 
 56 void add(ll &a,ll b)
 57 {
 58     a+=b;
 59     if(a>=MOD) a-=MOD;
 60     if(a<0) a+=MOD;
 61 }
 62 
 63 int lcp(int a,int b)
 64 {
 65     int res=0;
 66     per(i,m-1,0)
 67       if((a>>i&1)==(b>>i&1)) res++;
 68        else break;
 69     return res;
 70 }
 71 
 72 void solve()
 73 {
 74     m=read(),n=read();
 75     rep(i,1,n) l[i]=read(),r[i]=read();
 76     rep(v,0,1)
 77      rep(j,0,m+1)
 78       rep(k,0,m+1) s[v][j][k]=dp[v][j][k]=0;
 79     int v=0;
 80     dp[v][0][m]=s[v][0][m]=1;
 81     rep(i,0,m)
 82      per(j,m,0)
 83      {
 84          add(s[v][i][j],s[v][i][j+1]);
 85          if(i)
 86          {
 87             add(s[v][i][j],s[v][i-1][j]);
 88             add(s[v][i][j],-s[v][i-1][j+1]);
 89          }
 90      }
 91      int A,B,C,D;
 92      rep(i,1,n)
 93      {
 94          v^=1;
 95          rep(j,0,m)
 96           rep(k,0,m) s[v][j][k]=dp[v][j][k]=0;
 97          rep(j,l[i],r[i])
 98          {
 99             A=lcp(j,l[i]),B=lcp(j,r[i]);
100             if(i<n) C=lcp(j,l[i+1]);
101              else C=0;
102             if(i>1) D=lcp(j,r[i-1]);
103              else D=0;
104             dp[v][C][B]=(dp[v][C][B]+s[v^1][A][D]*j)%MOD;
105          }
106          rep(i,0,m)
107           per(j,m,0)
108           {
109                 s[v][i][j]=dp[v][i][j];
110                 add(s[v][i][j],s[v][i][j+1]);
111                 if(i)
112                 {
113                     add(s[v][i][j],s[v][i-1][j]);
114                     add(s[v][i][j],-s[v][i-1][j+1]);
115                 }
116           }
117      }
118      printf("%I64d\n",s[v][m][0]);
119 }
120 
121 int main()
122 {
123     //freopen("1.in","r",stdin);
124     //freopen("1.out","w",stdout);
125     int cas=read();
126     while(cas--) solve();
127     return 0;
128 }