UVa 12208 How Many Ones Needed? (组合数学)

12208 - How Many Ones Needed?

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=onlinejudge&Itemid=99999999&category=244&page=show_problem&problem=3360

水。

完整代码：

/*0.035s*/

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

ll c[31][31], sum;
int i, j, cnt;

void init()
{
    for (i = 0; i < 31; ++i)
        c[i][0] = c[i][i] = 1;
    for (i = 2; i < 31; ++i)
        for (j = 1; j < i; ++j)
            c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
}

inline ll calc(int n, bool yeah)
{
    sum = cnt = 0;
    for (i = 31; i >= 0; --i)
    {
        if (n & (1 << i))
        {
            for (j = 0; j <= i; ++j)
                sum += c[i][j] * (j + cnt);
            ++cnt;
        }
    }
    if (yeah) sum += cnt;///把数n也算上
    return sum;
}

int main()
{
    init();
    int a, b, cas = 0;
    ll ans;
    while (scanf("%d%d", &a, &b), a || b)
        printf("Case %d: %lld\n", ++cas, calc(b, true) - calc(a, false));
    return 0;
}

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