Codeforces Round #230 (Div. 2

连涨六次RATING居然还是不够紫T_T，不过确实分数不够（明明每次差一点！！！）

A.Nineteen

水题，统计每个字母出现次数，然后ans--验证

不知道哪还有BUG 还是有人FST

#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cctype>#include<cmath>#include<ctime>#include<utility>#define M0(x) memset(x, 0, sizeof(x))#define Inf 0x7fffffffusing namespace std;int main(){    char s[110];    scanf("%s",&s);    int i,j;    int len=0;    for(i=0;s[i]!='\0';i++)        len++;    int g[100];    memset(g,0,sizeof(g));    for(i=0;i<len;i++)    {        if(s[i]=='n')   g[1]++;        if(s[i]=='i')   g[2]++;        if(s[i]=='e')   g[3]++;        if(s[i]=='t')   g[4]++;    }    int ans=0;    for(i=100;i>=0;i--)    {        if( i*2+1<=g[1] && i<=g[2] && i*3<=g[3] && i<=g[4] )        {            ans=i;break;        }    }    printf("%d",ans);    return 0;}

B.Three matrices

对角分别为原数和0，然后比较对称位置差值，A为平均值

#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cctype>#include<cmath>#include<ctime>#include<utility>#define M0(x) memset(x, 0, sizeof(x))#define Inf 0x7fffffffusing namespace std;int main(){    int i,j;    double w[200][200];    double a[200][200],b[200][200];    memset(a,0.0,sizeof(a));    memset(b,0.0,sizeof(b));    int n;    scanf("%d",&n);    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)        {            scanf("%lf",&w[i][j]);        }    double p,q;    for(i=1;i<=n;i++)    {        for(j=i;j<=n;j++)        {            if(i==j)            {                a[i][j]=w[i][j];                b[i][j]=0;                continue;            }            p=(w[i][j]-w[j][i])/2;            b[i][j]=p;b[j][i]=0-p;            a[i][j]=w[i][j]-p;            a[j][i]=w[i][j]-p;        }    }    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)        {            printf("%lf ",a[i][j]);        }        printf("\n");    }    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)        {            printf("%lf ",b[i][j]);        }        printf("\n");    }    return 0;}

C.Blocked Points

定义二维平面上的两个点AB是4-连通
当且仅当两个点的几何距离是1且两个点都不是障碍物的点，
或者存在点C使得AC、BC都是4-连通
一开始整个二维平面上没有障碍物，设和原点距离不超过n的点为特殊点，
至少需要把多少个点变成障碍才能使特殊点和非特殊点之间都不是4-连通

其实就是把圆内点和圆外点分开，做一个1/8 或1/4圆弧的遍历，算出x+1后高的差，即为该x除需处理的点个数

0要特判一下。

之前每个x写的一次算出，结果精度不够T-T。所以还是拿别人代码来了

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<cmath>using namespace std ;typedef long long LL ;LL N ;LL ans=0;int main(){LL i, j, tmp ;scanf("%I64d", &N) ;if(N == 0)ans = 1 ;else{for(i = 1; i <= N-1; i ++){tmp = sqrt(N*N-i*i)+1e-9 ;while(true){if(tmp > 0 &&                    (tmp*tmp+(i+1)*(i+1) > N*N || (tmp+1)*(tmp+1)+i*i > N*N))                    ans ++ ;else break ;tmp -- ;}}ans = ans*4+4 ;}printf("%I64d\n", ans) ;//system("pause") ;return 0 ;}

另外这题有公式，见图

D.Tower of Hanoi

带价值汉诺塔游戏，之前考虑把普通汉诺塔程序过程拿来累加，不过貌似还是得DP

#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cctype>#include<cmath>#include<ctime>#include<utility>using namespace std;int t[3][3];int dis[3][3];long long dp[50][3][3][3];long long dfs(int s,int i,int j,int k){    long long ans=0;    if(s==1)    {        return dis[i][k];    }    if(dp[s][i][j][k]) return dp[s][i][j][k];    long long cost1=dfs(s-1,i,k,j)+t[i][k]+dfs(s-1,j,i,k);    long long cost2=dfs(s-1,i,j,k)+t[i][j]+dfs(s-1,k,j,i)+t[j][k]+dfs(s-1,i,j,k);    return dp[s][i][j][k]=min(cost1,cost2);}int main(){    int n;    for(int i=0; i<3; ++i)        for(int j=0; j<3; ++j)            scanf("%d",&t[i][j]);    for(int i=0; i<3; ++i)        for(int j=0; j<3; ++j)            for(int k=0; k<3; ++k)                if(i!=j&&j!=k&&i!=k)                    dis[j][k]=min(t[j][k],t[j][i]+t[i][k]);    scanf("%d",&n);    printf("%I64d",dfs(n,0,1,2));}

E.

Yet Another Number Sequence?

会看这个题解的人应该不需要

反正我也没看，有时间再补