Codeforces Round #206 (Div. 2)

A：

题意：

　　给你两个数k,d，k表示有多少位数， d表示不断的对一个数的个数为求和，最后的得到的0-9的数

思路：

　　if (k == 1) 直接输出d,

　　if (k != 1) if (d == 0) 非法 else 输入d *10^(k - 1)

int main()
{
    int k,d;
    scanf("%d%d",&k,&d);
    if (k == 1) printf("%d
",d);
    else
    {
        if (d == 0) printf("No solution
");
        else
        {
            string ans = "1";
            for (int i = 1; i < k - 1; ++i) ans += "0";
            ans += (d - 1) + '0';
            cout << ans << endl;
        }
    }
    return 0;
}
View Code

B：

题意：

　　有两种交通工具非别为n个bus,m个trolleys 。我们有四种票可以选择c1，c2,c3,c4分别描述如下

  1. A ticket for one ride on some bus or trolley. It costs c1 burles;

  2. A ticket for an unlimited number of rides on some bus or on some trolley. It costs c2 burles;

  3. A ticket for an unlimited number of rides on all buses or all trolleys. It costs c3 burles;

  4. A ticket for an unlimited number of rides on all buses and trolleys. It costs c4 burles.

给出每种交通工具使用的次数a[] 表示对应的bus使用的次数， b[]表示trolleys  使用的次数，问如何选票是的最后费用最少。

思路：

　　首先一个c4和2个c3是确定的，然后我们再看有一个c3的话，然后对另一中交通工具进行c1和c2的选择，然后我们计算a[i](/b[i])用c1省还是用c2省就可以了。 如果只有c1,c2的话同样计算。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define M 5007
#define N 1007
using namespace std;

const ll mod = 1000000009;
const ll inf = (1LL<<53);

int c1,c2,c3,c4;
int a[N],b[N];
int n,m;

int main()
{
    scanf("%d%d%d%d",&c1,&c2,&c3,&c4);
    scanf("%d%d",&n,&m);
    for (int i = 0; i < n; ++i) scanf("%d",&a[i]);
    for (int i = 0; i < m; ++i) scanf("%d",&b[i]);

    int ans = min(c4,2*c3);

    //1c3
    int s = c3;
    for (int i = 0; i < n; ++i)
    {
        int v1 = c1*a[i];
        int v2 = c2;
        s += min(v1,v2);
        if (s > ans) break;
    }
    ans = min(ans, s);
    s = c3;
    for (int i = 0; i < m; ++i)
    {
        int v1 = c1*b[i];
        int v2 = c2;
        s += min(v1,v2);
        if (s > ans) break;
    }
    ans = min(ans, s);

    s = 0;
    for (int i = 0; i < n; ++i)
    {
        int v1 = c1*a[i];
        int v2 = c2;
        s += min(v1,v2);
        if (s > ans) break;
    }
    for (int i = 0; i < m; ++i)
    {
        int v1 = c1*b[i];
        int v2 = c2;
        s += min(v1,v2);
        if (s > ans) break;
    }
    ans = min(ans, s);
    printf("%d
",ans);
    return 0;
}
View Code

C：

题意：

　　给你n个数a[]，我们可以从左边取，也可以从右边取数。从左边取一个数的的代价为a[i]*L,如果连续从左边取得花，每次还要加上一个QL， 从右边取同左边取得操作一样 R,QR。问如何去才能取得最小值？

思路：

　　不论如何取，总会有一个结束的点我们枚举结束的点，然后计算左边取了多少个，右边取了多少个，然后哪边多，哪边就加上多出那部分*Qx, x{L,R}即可

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define M 5007
#define N 100007
using namespace std;

const ll mod = 1000000009;
const int inf = 0x7fffffff;

int n,l,r,Ql,Qr;
int w[N];
int sumL[N],sumR[N];

int main()
{
    scanf("%d%d%d%d%d",&n,&l,&r,&Ql,&Qr);
    for (int i = 1; i <= n; ++i) scanf("%d",&w[i]);
    sumL[0] = 0;
    for (int i = 1; i <= n; ++i) sumL[i] = sumL[i - 1] + w[i]*l;
    sumR[0] = 0;
    for (int i = n, j = 1; i >= 1 && j <= n; --i,j++) sumR[j] = sumR[j - 1] + w[i]*r;

//    for (int i = 1; i <= n; ++i) printf("%d ",sumL[i]);
//    printf("
");
//    for (int i = 1; i <= n; ++i) printf("%d ",sumR[i]);
//     printf("
");
    int ans = min(sumL[n] + (n - 1)*Ql, sumR[n] + (n - 1)*Qr);
    for (int i = 1; i <= n - 1; ++i)
    {
        int x,y;
        x = i; y = n - i;
        if (x > y)
        {
            int tmp = sumL[x] + (x - y - 1)*Ql + sumR[y];
            ans = min(ans, tmp);
        } else if (x == y){
            int tmp = sumL[x] + sumR[y];
            ans = min(ans, tmp);
        }
        else {
            int tmp  =sumR[y] + (y - x - 1)*Qr + sumL[x];
           // printf("%d %d
",i,tmp);
            ans = min(ans, tmp);
        }
    }
    printf("%d
",ans);
    return 0;
}
View Code

D:

题意：

　　给你一个n*n的只含有英文小写字母的矩阵，然后我们从s[0][0]出发到目的地s[n - 1][n - 1]规定只能向下，向右走。每条路径对应了一个字符串，如果该字符串含有a的数量大于b的数量，1赢，anum == bnum 平局， anum < bnum 2赢。 他们轮流选择下一步该走哪一个。问最后谁赢？

思路：

　　首先1如果选择的话，肯定是选择a比较多的路线； 2如果选择的话肯定的a少b多的路线。 然后dp一下就好了。

By E_star, contest: Codeforces Round #206 (Div. 2), problem: (D) Game with Strings, Accepted, #
 //#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define keyTree (chd[chd[root][1]][0])
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define M 100007
#define N 100007

using namespace std;

int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};

const int inf = 0x7f7f7f7f;
const int mod = 1000000007;
const double eps = 1e-8;
const int R = 100007;


char s[22][22];
int n;
int dp[42][1<<20];
int a[42][26];
int ans;
int DP(int step,int state)
{
    if (dp[step][state] != 1e9) return dp[step][state];
    int add = 0;
    for (int i = 0; i < n; ++i)
    {
        if (state & (1<<i))
        {
            if (s[i][step - i] == 'a') add++;
            else if (s[i][step - i] == 'b') add--;
            break;
        }
    }
    if (step == 2*n - 2) return dp[step][state] = add;
    if (step % 2 == 0)
    {
        for (int i = 0; i < 26; ++i)
        {
            int next = a[step + 1][i] & (state | (state << 1));
            if (next == 0) continue;
            dp[step][state] = min(dp[step][state], DP(step + 1, next) + add);
        }
    } else {
        dp[step][state] = -1e9;
        for (int i = 0; i < 26; ++i)
        {
            int next = a[step + 1][i] & (state | (state << 1));
            if (next == 0) continue;
            dp[step][state] = max(dp[step][state], DP(step + 1, next) + add);
        }
    }
    return dp[step][state];
}
int main()
{
    scanf("%d",&n);
    for (int i = 0; i < n; ++i) scanf("%s", s[i]);
    CL(a,0);
    for (int i = 0; i < 2*n - 1; ++i)
    {
        for (int j = 0; j < n; ++j)
        {
            if (i - j < 0 || i - j >= n) continue;
            a[i][s[j][i - j] - 'a'] |= (1<<j);
        }
    }
    for (int i = 0; i < 2*n - 1; ++i)
        for (int j = 0; j < (1<<20); ++j)
            dp[i][j] = 1e9;
    int ans = DP(0,1);
    if (ans == 0) printf("DRAW
");
    else if (ans > 0) printf("FIRST
");
    else  printf("SECOND
");
    return 0;
}
View Code

E:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>


#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll __int64
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define keyTree (chd[chd[root][1]][0])
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define M 100007
#define N 300007

using namespace std;

int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};

const double inf = 100000000000.0;
const int mod = 1000000007;

int n,k;
int a[N];

int main()
{
    scanf("%d%d",&n,&k);
    for (int i = 0; i < n; ++i) scanf("%d",&a[i]);
    sort(a, a + n);
    if (a[0] <= k)
    {
        printf("%d
",a[0]);
    } else {
        int ans = a[0];
        while (1)
        {
            int i;
            for (i = 0; i < n; ++i)
            {
                if (a[i]%ans > k)
                {
                    while (a[i]%ans > k) ans--;
                    break;
                }
            }
            if (i == n) break;
        }

        printf("%d
",ans);
    }
    return 0;
}
View Code
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原文地址：https://www.cnblogs.com/E-star/p/3370905.html