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This one is marked as "Advanced".. i don‘t tink so, not that hard if you can visualize all the bits from a to b. Two key points here:
1. Say both a and b are positive, bits form an interesting pattern vertically - all bit[i] of all numbers
2. negative numbers, count(a) = 32 - count(1 - a)
The above observation leads to this code:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; long long _calc(long long a, long long b) { int bb = b; int pc = 1, sz = 2; long long ttl = 0; while (bb) { int sa = a / sz; int sb = b / sz; long long cnt = (long long)(sb - sa + 1) * (long long)pc; int ra = a % sz; if (ra > pc) cnt -= std::max(ra - pc, 0); int rb = b % sz; if (rb < sz - 1) cnt -= std::min(pc, sz - rb - 1); ttl += cnt; // Move on bb >>= 1; pc *= 2; sz *= 2; } return ttl; } long long calc(long long a, long long b) { long long cnt = 0; // negative int na, nb; if (a < 0) { na = a; nb = std::min((long long)-1, b); long long aa = -(nb + 1); long long bb = -(na + 1); long long ret = _calc(aa, bb); cnt = (long long)(bb - aa + 1) * 32 - ret; } long long pa, pb; if (b >= 0) { pa = std::max(a, (long long)0); pb = b; cnt += _calc(pa, pb); } return cnt; } int main() { int t; cin >> t; while (t--) { long long a, b; cin >> a >> b; auto r = calc(a, b); cout << r << endl; } return 0; }
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原文地址：http://www.cnblogs.com/tonix/p/4585077.html
