二進制中1的個數

#include<bits/stdc++.h> typedef long long ll; const ll MO = 1000000007; int T,bin[63],lens; ll n,ans,power[63],pre[63],nxt[63]; ll read(){ll x;scanf("%lld",&x);return x;} int main() { scanf("%d",&T), power[0] = 1; for (int i=1; i<=61; i++) power[i] = (power[i-1]<<1)%MO; for(int cas=1;cas<=T;cas++) { n = read(), ans = lens = 0; for (ll x=n; x; x>>=1) bin[++lens] = x&1; pre[0] = nxt[lens+1] = 0; for (int i=1; i<=lens; i++) pre[i] = (pre[i-1]+power[i-1]*bin[i])%MO; for (int i=lens; i>=1; i--) nxt[i] = ((nxt[i+1]<<1)+bin[i])%MO; for (int i=1; i<=lens; i++) { if (bin[i]) pre[i-1]++,ans = (ans+pre[i-1])%MO; ans = (ans+nxt[i+1]*power[i-1]%MO)%MO; } // ans爲1到n中i的二進制1的個數 //std::cout<<ans<<'\n';continue; ans = (n+1ll)%MO*ans%MO; for (int i=1; i<=lens; i++) { if (bin[i]) ans = (ans+MO-pre[i-1]*pre[i-1]%MO)%MO; ans = (ans-nxt[i+1]*power[i-1]%MO*power[i-1]%MO+MO)%MO; } // ans-[i,j]前綴1的個數 printf("Case #%d: %lld\n",cas,ans); } return 0; } // for(ll i=1;i<=n;i<<=1,++len) // if(n&i)ans+=n%i+1+(n>>len+1)<<len; // else ans+=(n>>len+1)<<len; // i屬於 [1,n]內i 的二進制下1的個數 總合