CF1245F Daniel and Spring Cleaning
题意：
给定l，r，求∑a=lr∑b=lr[a+b=a⊕b]\sum_{a=l}^{r}\sum_{b=l}^{r}[a+b=a⊕b]∑a=lr∑b=lr[a+b=a⊕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
#elsestartTime = clock ();freopen("data.in", "r", stdin);
#endif
}
void Time_test()
{
#ifdef ONLINE_JUDGE
#elseendTime= 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();
}
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