HihoCoder 1527 题解

转载请注明出处，http://blog.youkuaiyun.com/Bule_Zst/article/details/77116239

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题目：

When we are coding, we can replace $x × a$ by (x<<$a_0$) op1 (x<<$a_1$) op2 (x<<$a_2$) … opn (x<<$a_n$) to make the program faster where opi is + or -.

For example, you can replace x × 15 by (x<<4)-(x<<0).

In this problem we assume that operator‘+’,‘-’and ‘<<’will cost 1 unit of time and operator‘*’will cost infinity units of time. For example, (x<<4)-(x<<1) will cost 3 units of time.

Given a, what’s the minimum units of time does it take to compute the value of $(x × a)$?

Input

One ‘01’ string S : the binary representation of a (from the most significant digit to the least significant digit)

1 ≤ the length of string $S ≤ 10^6$

a > 0

Output

An integer : the answer

Sameple Input

1111

Sameple Output

3

题目大意：

任何一个整数都可以用一系列2的幂加减得到，加、减和幂次运算耗时为1，给定一个整数（以二进制字符串形式给出），问最短耗时

方法：

1.二幂拆分问题，详见这篇博客

直接给出代码：

// @Team    : nupt2017team12
// @Author  : Zst
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
#define LL long long
#define MOD 1000000007
#define CLR(a,x) memset(a,x,sizeof(a))
#define INF 0x3f3f3f3f
#define pb push_back
#define FOR(i,a,b) for( int i = ( a ); i <= ( b ); ++i )

const int N = 1e6;

int len;
char str[N];

int u, v;


int main()
{
    // freopen( "E.txt", "r", stdin );
    scanf( "%s", str );
    len = strlen( str );
    int index = -1;
    for( int i = len-1; i >= 0; i-- ) {
        if( str[i] == '1' ) {
            index = i;
            break;
        }
    }
    u = v = 1;
    for( int i = index-1; i >= 0; i-- ) {
        if( str[i] == '1' ) {
            u = min( u, v ) + 1;
        } else {
            cout<<u<<" "<<v<<endl;
            v = min( u, v ) + 1;
            cout<<v<<endl;
        }
    }
    printf( "%d\n", 2 * u - 1 );

    return 0;
}

2.动态DP（其实第一种方法，本质上也是DP）

先将给的字符串倒置，然后从左往右扫

• 如果遇到‘0’则ans[i] = ans[i-1]
• 如果遇到‘1’则有两种可能，ans[i]取两种可能中的较小值
  
  • 第一种，ans[i] = ans[i-1] + 1，直接加上扫到的1
  • 第二种，ans[i] = ans[j-1] + 2 - ( zero[i] - zero[j-1] )
    
    2表示，在str[j-1]的基础上，先加$2^{i+1}$，再减去$2^{j}$，这样，原先的x就会变成一个 在二进制表示下，从j位到i位全是1的新数。（i、j从0开始数）（str[n]表示前n位二进制串组成的数）
    
    如$x = (100110)_2$，则$(10)_2 + 2^{5+1} - 2^{2} = (111110)_2$
    
    zero[n]表示二进制串前n位中，0的个数。减去( zero[i] - zero[j-1] )，就是减去原先应该是0的位。

代码：

// @Team    : nupt2017team12
// @Author  : Zst
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
#define LL long long
#define MOD 1000000007
#define CLR(a,x) memset(a,x,sizeof(a))
#define INF 0x3f3f3f3f
#define pb push_back
#define FOR(i,a,b) for( int i = ( a ); i <= ( b ); ++i )

const int N = 1e6+7;
char str[N];
int len;
int zero[N];
int ans[N];


int main()
{
    // freopen( "E.txt", "r", stdin );
    while( scanf( "%s", str ) != EOF ) {
        CLR( zero, 0 );
        CLR( ans, 0 );

        len = strlen( str );
        reverse( str, str+len );
        zero[0] = ( str[0] == '0' );
        FOR( i, 1, len-1 ) {
            zero[i] = zero[i-1] + ( str[i] == '0' );
        }
        int j = 0;
        ans[0] = ( str[0] == '1' );
        FOR( i, 1, len-1 ) {
            if( str[i] == '0' ) {
                ans[i] = ans[i-1];
            } else {
                ans[i] = ans[i-1]+1;
                if( j == -1 ) {
                    ans[i] = min( ans[i], 2 - zero[i] );
                } else {
                    ans[i] = min( ans[i], ans[j-1] + 2 + ( zero[i] - zero[j-1] ) );
                }
            }
            // 对于i+1位更新j
            if( j == -1 ) {
                if( 2 - zero[i+1] >= ans[i-1] + 2 + ( zero[i+1] - zero[i-1] ) ) 
                    j = i;
            } else {
                if( ans[j-1] + 2 + ( zero[i+1] - zero[j-1] ) >= ans[i-1] + 2 + ( zero[i+1] - zero[i-1] ) )
                    j = i;
            }
            cout<<j<<endl;
        }
        printf( "%d\n", ans[len-1] * 2 - 1 );
    }

    return 0;
}

参考博客：http://blog.youkuaiyun.com/werkeytom_ftd/article/details/73740918