UVa 12208 how Many ones Needed? (Combinatorial mathematics)

Source: Internet
Author: User
Tags cas time limit

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


Complete code:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
ll c[31][31], sum;
int I, j, CNT;
void init ()
    (i = 0; i < ++i)
        c[i][0] = c[i][i] = 1;
    for (i = 2; i < n ++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 = to i >= 0;-i)
        if (n & (1 << i))
            (j = 0; J <= i; ++j)
                sum = C I [j] * (j + CNT);
    if (yeah) sum + = cnt;///The number n is also counted as return
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;

Author: csdn Blog Synapse7

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