Gym102900C.Sum of Log（数位DP+剪枝）

数位DP+剪枝，本题卡$10^3的数位DP。理解还是不够深刻。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=33;
const int mod=1e9+7;
//定义f(i,j,k)
//到第i位位置，x是否卡死，y是否卡死，前面是否全是0的贡献。 
ll f[maxn][2][2][2];
 
int a[100],b[maxn],len1,len2;
ll L,R;
ll dfs (int pos,int limit1,int limit2,int st) {
	//printf("%d %d
",j,k);
	if (pos<=0) {
		return 1;
	}
	if (f[pos][limit1][limit2][st]!=-1)  return f[pos][limit1][limit2][st]%mod;
	ll ans=0;
	ll k1=limit1?a[pos]:1;
	ll k2=limit2?b[pos]:1;
	for (int x=0;x<=k1;x++) {
		for (int y=0;y<=k2;y++) {
			if (x&&y) continue;
			int tt;
			if (st&&(x||y)) tt=pos;
			else tt=1;
			ans+=1ll*tt*dfs(pos-1,limit1&&x==k1,limit2&&y==k2,st&&!x&&!y)%mod;
			ans%=mod;
		} 
	}
	f[pos][limit1][limit2][st]=ans;
	return ans;
} 
ll solve (ll L,ll R) {
	len1=0,len2=0;
	while (L) {
		a[++len1]=L%2;
		L/=2;
	}
	while (R) {
		b[++len2]=R%2;
		R/=2;
	}
	while (len1<len2) a[++len1]=0;
	while (len2<len1) b[++len2]=0;
	memset(f,-1,sizeof(f));
	return dfs(len1,1,1,1);
}
int main () {
	int _;
	cin>>_;
	while (_--) {
		ll L,R;
		scanf("%lld%lld",&L,&R);
		cout<<solve(L,R)-1<<'
';
	}
}