Codeforces 106 DIV2 ACD

B表示完全看不懂。。就不弄了。。

E字符串先不管了。到时候系统学下字符串再处理

A

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
int src[13];
int main()
{
    int K;
    scanf("%d",&K);
    for (int i = 0; i < 12; i++) scanf("%d",&src[i]);
    sort(src,src+12);
    int ans  = 0 ,sum  = 0, cas = 11;
    while (true)
    {
        if (sum >= K) break;
        if (ans == 13) break;
        sum += src[cas];
        cas--;
        ans++;
    }
    if (ans == 13) puts("-1");
    else printf("%d\n",ans);
    return 0;
}

C

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
#define MAXN 100005
int N;
struct node
{
    int id;
    int val;
    friend bool operator < (const node & a,const node &b)
    {
        return a.val < b.val;
    }
}src[MAXN];
LL sum[MAXN];
node stl[MAXN],str[MAXN];
int main()
{
    scanf("%d",&N);
    sum[0] = 0;
    for (int i = 1; i <= N; i++)
    {
        scanf("%d",&src[i].val);
        src[i].id = i ;
        sum[i] = sum[i - 1] + src[i].val;
    }
    sort(src + 1, src + 1 + N);
    int topl = 1 ,topr = 1;
    if (N % 2 == 0)
    {
       for (int i = 1; i  <= N; i++)
        {
            if (i % 2 == 0)  stl[topl++] =src[i];
            else str[topr++] = src[i];
        }
        printf("%d\n",N / 2);
        for (int i = 1; i < topl; i++) printf("%d ",stl[i].id);
        putchar(‘\n‘);
        printf("%d\n",N / 2);
        for (int i = 1; i < topr; i++) printf("%d ",str[i].id);
        putchar(‘\n‘);
    }
    else
    {
        for (int i = 1; i <= N; i += 2)
            stl[topl++] =src[i];
        for (int i = 2; i <= N; i += 2)
            str[topr++] = src[i];
        printf("%d\n",N / 2 + 1);
        for (int i = 1; i < topl; i++ ) printf("%d ",stl[i].id);
        putchar(‘\n‘);
        printf("%d\n", N / 2);
        for (int i = 1; i < topr; i++) printf("%d ",str[i].id);
        putchar(‘\n‘);
    }
    return 0;
}

D

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
#define MOD 1000000007
char input[720];
struct node
{
        char res;
        int id;
}src[720];
int G[720];
node sta[720];
LL dp[720][720][4][4];
void calcu(int l ,int r)
{
        if (l >= r) return ;
        if (l  + 1 == r)
        {
                dp[l][r][0][1] = 1;
                dp[l][r][1][0] = 1;
                dp[l][r][2][0] = 1;
                dp[l][r][0][2] = 1;
        }
        if (G[l] == r)
        {
                calcu(l + 1,r - 1);
                for (int i = 0 ; i < 3; i++)
                        for (int j = 0; j < 3; j++)
                {
                        if (j != 1) dp[l][r][0][1] = (dp[l][r][0][1] + dp[l + 1][r - 1][i][j]) % MOD;
                        if (j != 2) dp[l][r][0][2] = (dp[l][r][0][2] + dp[l + 1][r - 1][i][j]) % MOD;
                        if (i != 1) dp[l][r][1][0] = (dp[l][r][1][0] + dp[l + 1][r - 1][i][j]) % MOD;
                        if (i != 2) dp[l][r][2][0] = (dp[l][r][2][0] + dp[l + 1][r - 1][i][j]) % MOD;
                }
                return ;
        }
        else
        {
                 int t = G[l];
                 calcu(l,t);
                 calcu(t + 1, r);
                 for (int i = 0 ; i < 3; i++)
                        for (int j = 0 ; j < 3; j++)
                           for (int m = 0; m < 3; m++)
                               for (int n = 0 ; n < 3; n++)
                 {
                         if (!( (m == 1 && n == 1) || (m == 2 && n == 2)))
                             dp[l][r][i][j] = (dp[l][r][i][j] + (dp[l][t][i][m] * dp[t + 1][r][n][j]) % MOD) % MOD;
                 }
        }
}
int main()
{
        while (scanf("%s",input + 1) != EOF)
        {
                int len = strlen(input + 1);
                int top = 1;
                for (int i = 1; i <= len; i++)
                {
                        if (input[i] == ‘(‘)
                        {
                                node tmp;
                                tmp.id = i;
                                tmp.res = input[i];
                                sta[top++] = tmp;
                        }
                        else
                        {
                                if (sta[top - 1].res == ‘(‘)
                                {
                                        G[i] = sta[top - 1].id;
                                        G[sta[top - 1].id] = i;
                                        top--;
                                }
                                else
                                {
                                        node tmp;
                                        tmp.id = i;
                                        tmp.res = input[i];
                                        sta[top++] = tmp;
                                }
                        }
                }
                memset(dp,0,sizeof(dp));
                calcu(1,len);
                LL ans = 0;
                for (int i = 0 ; i < 3; i++)
                        for (int j = 0 ; j < 3; j++)
                        ans = (ans + dp[1][len][i][j]) % MOD;
                printf("%I64d\n",ans % MOD);
        }
        return 0;
}