模板 - 数位dp

#include
using namespace std;
#define ll long long

int a[40];
ll dp[40][40][40];
ll dfs(int pos, int s1, int s2, bool lead, bool limit) {
    if(pos == -1) {
        if(s1 >= s2)
            return 1;
        else
            return 0;
    }
    if(!limit && !lead && dp[pos][s1][s2] != -1)
        return dp[pos][s1][s2];
    int up = limit ? a[pos] : 1;
    ll ans = 0;
    for(int i = 0; i <= up; i++) {
        int ns1 = s1 + ((!lead) && i == 0);
        int ns2 = s2 + (i == 1);
        ans += dfs(pos - 1, ns1, ns2, lead && i == 0, limit && i == a[pos]);
    }
    if(!limit && !lead)
        dp[pos][s1][s2] = ans;
    return ans;
}

ll solve(ll x) {
    if(x < 0)
        return 0;
    int pos = 0;
    while(x) {
        a[pos++] = x % 2;
        x /= 2;
    }
    return dfs(pos - 1, 0, 0, true, true);
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif
    memset(dp, -1, sizeof(dp));

    ll le, ri;
    while(~scanf("%lld%lld", &le, &ri)) {
        printf("%lld\n", solve(ri) - solve(le - 1));
    }
}

转载于:https://www.cnblogs.com/Inko/p/11600358.html