CF1245F Daniel and Spring Cleaning(等会了更新)

Posted 2021-10-19 Jozky86

CF1245F Daniel and Spring Cleaning

题意：

给定l，r，求${\sum }_{a=l}^r{\sum }_{b=l}^r[a+b=a\oplus b]$

题解：

对于这个式子，只有当a和b都不为0时成立，也就是我们不求
对于这个式子有f(x,y)=f(y,x)
考虑差分，f(r,r)-f(l-1,r)-f(r,l-1)+f(l-1,l-1)
答案为：f(r,r)-2f(l-1,r)+f(l-1,l-1)
数位dp转移即可，数位dp还是dfs的转移好写

代码：

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef ONLINE_JUDGE
#else
    startTime = clock ();
    freopen("data.in", "r", stdin);
#endif
}
void Time_test()
{
#ifdef ONLINE_JUDGE
#else
    endTime= clock();
    printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn=50;
ll dp[maxn][2][2][2];
int b[maxn],a[maxn];
ll dfs(int len,int flag1,int flag2,int flag3){
	if(!len){
		return flag3;
	}
	if(dp[len][flag1][flag2][flag3]!=-1)return dp[len][flag1][flag2][flag3];
	ll res=0;
	int up1=flag1?a[len]:1;
	int up2=flag2?b[len]:1;
	for(int i=0;i<=up1;i++){
		for(int j=0;j<=up2;j++){
			bool F=flag3;
			if((i&j)!=0)F=0;
			res+=dfs(len-1,(i==up1)&&flag1,(j==up2)&&flag2,F);
		}
	}
	return dp[len][flag1][flag2][flag3]=res;
}

ll solve(ll x,ll y){
	if(x<0||y<0)return 0;
	int num1=0,num2=0; 
	memset(dp,-1,sizeof(dp));
	for(int i=0;i<=40;i++){
		a[i]=0;
		b[i]=0;
	}
	while(x){
		a[++num1]=x%2;
		x/=2;
	}
	while(y){
		b[++num2]=y%2;
		y/=2;
	}
	ll ans=dfs(32,1,1,1);
	return ans;
}
int main()
{
    rd_test();
	int t;
	read(t);
	while(t--){
		ll l,r;
		read(l,r);
		cout<<solve(r,r)-2ll*solve(l-1,r)+solve(l-1,l-1)<<endl;
	}
	return 0;
    //Time_test();
}