How to solve the problem of # Greed?

The cabbage button only has two states: 1 and 0, and silly, silly ljw wants to turn a bunch of cabbage buttons into the target state. But ljw has a rounded body, and when you push a button, you accidentally touch the next two. We know that if a cabbage button is pressed, it changes its state, and we find the least number of operations ljw.

Input format
two lines, giving two equal length strings of 0 and 1(no longer than 5*10^6), representing the current and target states.

Output format
The minimum number of push-button operations that need to be performed, or output impossible if the transformation cannot be achieved.

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dayin781m 注册会员
2023-02-26 16:10

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thank you

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dd261591875 注册会员
2023-02-26 16:10

,a+ 1 ,b+ 1 ); int len= strlen (a+ 1 ); for ( int i= 1 ;i<=len;i++) { if (a[i]==b[i])a[i]=b[i]= 0 ; else a[i]=b[i]= 1 ; } if (len== 1 ) { if (a[ 1 ]== 0 ) printf ( "0\n" ); else printf ( "1\n" ); return 0 ; } if (len== 2 ) { if (a[ 1 ]!=a[ 2 ]) printf ( "impossible\n" ); else if (a[ 1 ]==a[ 2 ]) { if (a[ 1 ]== 0 ) printf ( "0\n" ); else printf ( "1\n" ); } return 0 ; } int n1= 1 ,n2= 0 ; a[ 1 ]^= 1 ;a[ 2 ]^= 1 ; for ( int i= 1 ;i<=len -1 ;i++) { if (a[i]== 1 ){a[i]^= 1 ;a[i+ 1 ]^= 1 ;a[i+ 2 ]^= 1 ;n1++;} if (b[i]== 1 ){b[i]^= 1 ;b[i+ 1 ]^= 1 ;b[i+ 2 ]^= 1 ;n2++;} } if (a[len]== 1 )n1= 999999999 ; if (b[len]== 1 )n2= 999999999 ; int ans= min (n1,n2); if (ans== 999999999 ) printf ( "impossible\n" ); else printf ( "%d\n" ,ans); return 0 ; } .

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daiyuchun 注册会员
2023-02-26 16:10

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please click accept

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Publish Time: 2023-02-26 16:10

Update Time: 2023-02-26 16:10