Codeforces Round #206 (Div. 2)-白红宇

Codeforces Round #206 (Div. 2)

阅读量：4966 次

发布时间：2019-06-12

本文共 7603 字，大约阅读时间需要 25 分钟。

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\n",d);    else    {        if (d == 0) printf("No solution\n");        else        {            string ans = "1";            for (int i = 1; i < k - 1; ++i) ans += "0";            ans += (d - 1) + '0';            cout << ans << endl;        }    }    return 0;}

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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

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

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

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

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。

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                    #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\n", 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 1007using 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\n",ans); return 0;}

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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

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                    #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\n", 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 100007using 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("\n");// for (int i = 1; i <= n; ++i) printf("%d ",sumR[i]);// printf("\n"); 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\n",i,tmp); ans = min(ans, tmp); } } printf("%d\n",ans); return 0;}

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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 
      
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                    #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\n", 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 100007using 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<
                    
                     = n) continue; a[i][s[j][i - j] - 'a'] |= (1<
                     
                       0) printf("FIRST\n"); else printf("SECOND\n"); return 0;}

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#include 
      
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            #include 
            
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                    #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\n", 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 300007using 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\n",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\n",ans); } return 0;}

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转载于:https://www.cnblogs.com/E-star/p/3370905.html

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