洛谷P1106 删数问题

如下。这可能是思路比较清晰的代码了。

为了各位阅读方便，我将变量名加长了。

谢谢阅读233

#include<bits/stdc++.h>

using namespace std;

char ints[250];

int main()

{

int n;

cin>>ints>>n;

int length;length = strlen(ints);

static int currents[100];

int currentCheckingDigit=1;

int ptr=0;

while (currentCheckingDigit<=length){

if (n!=0){

int min =10000;int rminat=0;//relative min at(numbers needed to be off)

for (int i=ptr;i<=ptr+n;i++){

if (ints[i]-'0'<min) {min=ints[i]-'0';rminat=i-ptr;}//find out the minium and where it is

}

for (int i=ptr;i<=ptr+rminat-1;i++){

ints[i]='X';//everything before ptr+rminat will be elimated

}

n-=rminat;

ptr+=rminat+1;

currentCheckingDigit++;

}

else{

currentCheckingDigit++;

}

}

for (int i=0;i<=length-1;i++){

if (ints[i]=='0'||ints[i]=='X') ints[i]='X';

if (ints[i]!='0'&&ints[i]!='X') break;

}

for (int i=0;i<=length-1;i++) if (ints[i]!='X')cout<<ints[i];

system("pause");

return 0;

}