from HihoCoder 1527 题解
题目：
When we are coding, we can replace by (x<<
) op1 (x<<
) op2 (x<<
) ... opn (x<<
) 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 ?
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
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，再减去
，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，则
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;
}
