Codeforces Round #145 (Div. 2, ACM-ICPC Rules)

A题 很简单，没什么好写的，直接上代码。

#include <stdio.h>
#include <string.h>
#include <cstdio>
#include <math.h>
#include <stdlib.h>
int main()
{
    int n;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    scanf("%d", &n);
    char s[1000];
    scanf("%s", s);
    int len = strlen(s);
    int j = len/2;
    for(int i = 0; i < len/2; i++)
    {
        while(j<len)
        {
            if(s[i]!=s[j])
                if(s[i]=='R'&&s[j]=='L')
                {
                    printf("%d %d\n",j+1, i+1);
                    j++;
                    break;
                }
            printf("%d %d\n", i+1, j+1);
            j++;
            break;
        }
    }


    return 0;
}

B题

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#define eps 1e-5
#define MAXN 55555
#define MAXM 111111
#define INF 1000000007
#define lson (pos<<1)
#define rson (pos<<1|1)
using namespace std;

struct node
{
    int v;int No;
}a[10000];
int cmp(node a, node b)
{
    return a.v>b.v;
}
int main()
{
    int n, k;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    scanf("%d%d", &n,&k);

    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i].v);
        a[i].No = i;
    }
    sort(a+1, a+1+n, cmp);
    int f[10000];
    memset(f, 0, sizeof(f));
    for(int i = 1; i <= k; i++)
    {
        f[a[i].No] = 1;
    }
    printf("%d\n", a[k].v);
    for(int i = 1; i <= n; i++)
    {
        if(f[i])
            printf("%d ", i);
    }
    printf("\n");

    return 0;
} 

C题     这个题我在每一个点都判断一下所需操作数

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
int a[100010];
struct node
{
    int da;
    int xiao;
}b[100010];
int main()
{
    int n;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    int up = 0, down = 0, zero=0;
    for(int i = 0; i < n; i++)
    {
        if(a[i]>0)
            up++;
        if(a[i]<0)
            down++;
        if(a[i]==0)
            zero++;
        b[i].da = up;
        b[i].xiao = down;
     //   printf("i=%d up=%d down=%d zero=%d\n", i, up, down,zero);
    }
    int m = 0x7ffffff;
    for(int i = 0; i < n-1; i++)
    {
        int tmp = b[i].da+down-b[i].xiao;
        if(m>tmp)
            m = tmp;
        //printf("i=%d up=%d down=%d tmp=%d m=%d\n", i, b[i].da, b[i].xiao,tmp,m);
    }
    printf("%d\n", m+zero);
    return 0;
}