lightoj - 1032 - Fast Bit Calculations - 数位dp

来源：互联网发布：eia数据今天数据编辑：程序博客网时间：2024/06/06 08:35

题意：将一个十进制数转化为二进制后，统计这一位是1，且下一位也是1的位数为这个数的权值，例如，6的二进制为110，第一1的下一位也是1,所以6的权值为1。给一个十进制数n，求0到n所有的十进制数的权值和。

题解：数位dp，可以参考算法合集之《浅谈数位类统计问题》，用树形图来理解就容易多了。

#include <bits/stdc++.h>//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define ll long long#define SZ(x) ((int)(x).size()) #define ALL(v) (v).begin(), (v).end()#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i) #define REP(i,a,n) for ( int i=a; i<int(n); i++ )#define FOR(i,a,n) for ( int i=n-1; i>= int(a);i-- )#define lson rt<<1, L, m#define rson rt<<1|1, m, Rtypedef pair<int, int> pii;typedef pair<ll, ll> pll;#define mp(x, y) make_pair(x, y)#define pb(x) push_back(x)#define fi first#define se second#define CLR(a, b) memset(a, b, sizeof(a))#define Min(a, b) a = min(a, b)#define Max(a, b) a = max(a, b)const int maxn = 50;int T;int kase;int n;int a[maxn];ll dp[maxn][2];ll cnt[maxn];void ini(){    REP(i, 1, maxn){        dp[i][1] = dp[i-1][0] + dp[i-1][1] + (1 << (i - 1));        dp[i][0] = dp[i-1][0] + dp[i-1][1];    }}ll solve(){    CLR(a, 0);    CLR(cnt, 0);    ll res = 0;    n ++;    int m = 0;    while(n){        a[m ++] = n % 2;        n /= 2;    }    REP(i, 0, m){        cnt[i+1] = cnt[i];        if(a[i]) res += dp[i][0], cnt[i+1] += (1 << i);        if(a[i] && a[i+1]) res += cnt[i];    }    return res;}int main(){#ifdef ac    freopen("in.txt","r",stdin);#endif    //freopen("out.txt","w",stdout);    scanf("%d", &T);    ini();    while(T--){        scanf("%d", &n);        printf("Case %d: %lld\n", ++ kase, solve());    }    return 0;}

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