CodeForces 738D Sea Battle

抽屉原理。

先统计最多有$sum$个船可以放，假设打了$sum-a$枪都没打中$a$个船中的任意一个，那么再打$1$枪必中。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0);
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c))
    {
        x = x * 10 + c - ‘0‘;
        c = getchar();
    }
}

int n,a,b,k;
char s[200010];
struct X
{
    int L;
    int cnt;
}t[200010];
int sz,cnt;
int sum;

int main()
{
    cin>>n>>a>>b>>k;
    cin>>s;

    for(int i=0;s[i];)
    {
        if(s[i]==‘0‘)
        {
            int j;
            for(j=i;s[j];j++)
            {
                if(s[j]==‘0‘) continue;
                break;
            }
            j--;

            t[sz].L=i;
            t[sz].cnt=j-i+1;
            sz++;
            i=j+1;
        }
        else i++;
    }

    for(int i=0;i<sz;i++) sum=sum+t[i].cnt/b;

    int now=0;

    printf("%d\n",sum-a+1);

    for(int i=0;i<sz;i++)
    {
        for(int j=1;j<=t[i].cnt/b;j++)
        {
            printf("%d ",t[i].L+j*b);
            now++;
            if(now==sum-a+1) break;
        }
        if(now==sum-a+1) break;
    }

    return 0;
}