杭电的集训队选拔赛，基本都是模拟题……

没什么算法之类的东西，做起来挺有感觉，俗称刷水题。

3343脑筋急转弯。。蚂蚁速度不为0就能走到。

3344看懂题意即可，水题，完全感觉不到这题原版的鸭梨。

#pragma warning(disable:4786)
#include "cstdlib"
#include "cctype"
#include "cstring"
#include "cstdio"
#include "cmath"
#include "algorithm"
#include "vector"
#include "string"
#include "iostream"
#include "sstream"
#include "set"
#include "queue"
#include "stack"
#include "fstream"
#include "iomanip"
#include "bitset"
#include "list"
#include "ctime"
using namespace std;
 
typedef __int64					LL;
typedef unsigned __int64		ULL;
typedef unsigned char			uchar;
#define CC(m,what)				memset(m,what,sizeof(m))
#define FOR(i,a,b)				for( i = (a); i < (b); ++i )
#define FF(i,a)					for( i = 0; i < (a); ++i )
#define FFD(i,a)				for( i = (a)-1; i >= 0; --i )
#define FORD(i,a,b)				for( i = (a)-1; i >= (b); --i )
#define LEFT(a)					((a)<<1)
#define RIGHT(a)				(((a)<<1)+1)
#define SZ(a)					((int)a.size())
#define PP(n,m,a)				puts("---");FF(i,n){FF(j,m)printf("%10d", a[i][j]);puts("");}
const double eps = 1e-11;
const double Pi = acos(-1.0);
const int    dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};

int		map[110][110];

void hexie() {
	int n, m, i, j, k, ix, iy;
	char x;
	bool mark;
	scanf("%d%d%*c", &n, &m);
	FF (i, n) {
		FF (j, m) {
			scanf("%c%*c", &x);
			if ( x == '_' ) map[i][j] = -1;
			else map[i][j] = x-'0';
		}
	}
	FF (i, n) {
		FF (j, m) {
			if ( map[i][j] != -1 ) {
				printf( j == m-1 ? ".......\n" : "....... ");
				continue;
			}
			ix = i+dir[1][0];
			iy = j+dir[1][1];
			if ( ix < 0 || iy < 0 || ix >= n || iy >= m || map[ix][iy] == -1 ) {
				mark = false;
				printf("XXX");
			} else {
				k = 0;
				mark = true;
				while ( map[ix][iy] != -1 ) {
					k += map[ix][iy];
					ix += dir[1][0];
					iy += dir[1][1];
					if ( ix < 0 || iy < 0 || ix >= n || iy >= m || map[ix][iy] == -1 ) {
						printf("%03d", k);
						break;
					}
				}
			}
			ix = i+dir[0][0];
			iy = j+dir[0][1];
			if ( ix < 0 || iy < 0 || ix >= n || iy >= m || map[ix][iy] == -1 ) {
				if ( !mark ) 
					printf(j == m-1 ? "XXXX\n" : "XXXX ");
				else 
					printf(j == m-1 ? "\\XXX\n" : "\\XXX ");
			} else {
				k = 0;
				while ( map[ix][iy] != -1 ) {
					k += map[ix][iy];
					ix += dir[0][0];
					iy += dir[0][1];
					if ( ix < 0 || iy < 0 || ix >= n || iy >= m || map[ix][iy] ==-1 ) {
						printf( j == m-1 ? "\\%03d\n" : "\\%03d ", k);
						break;
					}
				}
			}
		}
	}
}

int main() {
	int T, res, t;
	scanf("%d%*c", &T);
	while ( T-- ) {
		hexie();
		puts("");
	}
	return 0;
}

3345优先队列，如果要用队列的话要重复标记

#include "cstdlib"
#include "cctype"
#include "cstring"
#include "cstdio"
#include "cmath"
#include "algorithm"
#include "vector"
#include "string"
#include "iostream"
#include "sstream"
#include "set"
#include "queue"
#include "stack"
#include "fstream"
#include "iomanip"
#include "bitset"
#include "list"
#include "ctime"
using namespace std;

typedef unsigned char			uchar;
#define CC(m,what)				memset(m,what,sizeof(m))
#define FOR(i,a,b)				for( i = (a); i < (b); ++i )
#define FF(i,a)					for( i = 0; i < (a); ++i )
#define FFD(i,a)				for( i = (a)-1; i >= 0; --i )
#define FORD(i,a,b)				for( i = (a)-1; i >= (b); --i )
#define LEFT(a)					((a)<<1)a
#define RIGHT(a)				(((a)<<1)+1)
#define SZ(a)					((int)a.size())
#define P(n,a)					printf("---");FF(i,n){printf("%d:%4d\n", i,a[i]);printf("");}
#define PP(n,m,a)				printf("---");FF(i,n){FF(j,m)printf("%10d", a[i][j]);printf("");}
#define PPc(n,m,a)				printf("---");FF(i,n){FF(j,m)printf("%c", a[i][j]);printf("");}
const double eps = 1e-11;
const double Pi = acos(-1.0);
const int	intmax = 2147483647;
const int	dir[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};

class node {
public:
	int x, y, value;
};

struct cmp {
	bool operator () (const node &a,const node &b){
		return a.value < b.value;
	}
};

int			n, m, sx, sy, v;
char		map[110][110];
bool		mark[110][110];

bool findE(node t) {
	int i;
	FF (i, 4 ) {
		node q = t;
		q.x += dir[i][0];
		q.y += dir[i][1];
		if ( q.x < 0 || q.y < 0 || q.x >= n || q.y >= m ) continue;
		if ( map[q.x][q.y] == 'E' ) return true;
	}
	return false;
}

void bfs() {
	priority_queue<node,vector<node>,cmp> Q;
	node p, q;
	int i, j, k;
	p.x = sx;
	p.y = sy;
	p.value = v;
	Q.push(p);
	mark[p.x][p.y] = true;

	while ( !Q.empty() ) {
		p = Q.top();
		Q.pop();
		FF (i, 4) {
			q = p;
			q.x += dir[i][0];
			q.y += dir[i][1];
			if ( q.x < 0 || q.y < 0 || q.x >= n || q.y >= m || mark[q.x][q.y] || map[q.x][q.y] == '#' || map[q.x][q.y] == 'E' ) continue;
			if ( map[q.x][q.y] == '.' || map[q.x][q.y] == 'P' ) {
				--q.value ;
			} else if ( map[q.x][q.y] == 'T' ) {
				q.value -= 2;
			} else if ( map[q.x][q.y] == 'R' ) {
				q.value -= 3;
			} 
			if ( q.value < 0 ) continue;
			if ( findE(q) ) q.value = 0;
			if ( map[q.x][q.y] != 'P' ) 
				map[q.x][q.y] = '*';
			mark[q.x][q.y] = true;
			Q.push(q);
		}
	}
}

void hexie() {
	scanf("%d%d%d%*c", &n, &m, &v);
	int i, j, k;
	FF (i, n) gets(map[i]);
	FF (i, n) FF (j, m) {
		if ( map[i][j] == 'Y' ) {
			sx = i;
			sy = j;
		}
	}
	CC(mark,false);
	bfs();
    FF (i, n) puts(map[i]);putchar('\n');
}

int main() {
	int T;
	scanf("%d%*c", &T);
	while ( T-- ) {
		hexie();
	}
	return 0;
}

3346水题，表示看的时候还用了个大数模板，结果发现不用也能过，按题意来就好了。

3347模拟，表示用stl的map可能简化很多代码，以下是没用map的代码，头文件在上面几个代码里有。

class node {
public:
    char s[1000];
    int value;
};

node t[1000], temp;
int  res, n;
char s[1000];

void play() {
    int i, j, k;
    int v;
    res = 0;
    bool mark = false;
    for ( i = 0; ; i += strlen(s)+1 ) {
        sscanf(temp.s+i, "%s", s);
//        cout << s << endl;
        if ( s[0] == '=' ) continue;
        else if ( s[0] == '?' ) return;
        else if ( s[0] == '+' ) mark = false;
        else if ( strlen(s) == 1 && s[0] == '-' ) mark = true;
        else {
            if ( sscanf(s, "%d", &v) == 0 ) {
                FFD (j, n) {
                    if ( strcmp(t[j].s, s) == 0 ) {
                        res += mark ? -t[j].value : t[j].value;
                        break;
                    }
                }
            } else {
                res += mark ? -v : v;
            }
        }
    }
}

void hexie() {
    int i;
    scanf("%d%*c", &n);
    FF (i, n-1) {
        scanf("%s = %d%*c", temp.s, &temp.value);
        t[i] = temp;
    }
    gets(temp.s);
    play();
    printf("%d\n", res);
}

int main() {
    int T;
    scanf("%d%*c", &T);
    while ( T-- ) {
        hexie();
    }
    return 0;
}

3348 技巧贪心，表示描述不清楚贪心策略，看代码好点，没优化。

int        a[5], p, most, mini, t;
int        value[5] = {1, 5, 10, 50, 100};

bool getmax() {
    int temp;
    if ( a[0] >= t ) {
        most += t;
        if ( t >= 0 ) 
            return true;
        else 
            return false;
    } else if ( a[0] + a[1]*5 >= t ) {
        temp = a[0];
        int k = (t-temp-1)/5+1;
        t -= k*5;
        most += k;
        return getmax();
    } else if ( a[0] + a[1]*5 + a[2]*10 >= t ) {
        temp = a[0] + a[1]*5;
        int k = (t-temp-1)/10+1;
        t -= 10*k;
        most += k;
        return getmax();
    } else if ( a[0] + a[1]*5 + a[2]*10 + a[3]*50 >= t ) {
        temp = a[0] + a[1]*5 + a[2]*10 ;
        int k = (t-temp-1)/50+1;
        t -= 50*k;
        most += k;
        return getmax();
    } else if ( a[0] + a[1]*5 + a[2]*10 + a[3]*50 + a[4]*100 >= t ) {
        temp = a[0] + a[1]*5 + a[2]*10 + a[3]*50;
        int k = (t-temp-1)/100+1;
        t -= 100*k;
        most += k;
        return getmax();
    }
    return false;
}

void getmin() {
    mini = 0;
    int t = p;
    int i;
    FFD (i, 5) {
        if ( t > a[i]*value[i] ) {
            t -= a[i]*value[i];
            mini += a[i];
        } else {
            mini += t/value[i];
            t = t%value[i];
        }
    }
}

void hexie() {
    int i, j, k;
    scanf("%d", &p);
    FF (i, 5) scanf("%d", a+i);
    most = mini = 0;
    t = p;
    if ( !getmax() ) {
        puts("-1 -1");
        return;
    }
    getmin();
    printf("%d %d\n", mini, most);
}

int main() {
    int T;
    scanf("%d%*c", &T);
    while ( T-- ) {
        hexie();
    }
    return 0;
}

3349数学题

int main()
{
    int t;
    double lo,a,b,sma,biao;
    cin>>t;
    while(t--) {
        cin>>lo>>a>>b;
        biao = sqrt(2.0)*lo;
        sma = a<b? a: b;
        double mid = sma*1.0/2;
        if(mid<=(biao*1.0/2)) printf("%.4lf\n",mid*mid);
        else if((mid>(biao*1.0)/2)&&(mid<biao*1.0)) 
            printf("%.4lf\n",lo*lo - (biao-mid)*(biao-mid));
        else printf("%.4lf\n",lo*lo);
        
    }
    return 0;
}

 3350按题意模拟

char    st[1010];

int dfs(int s, int e, int &num) {
    int i = s, t1, t2, res, j, l, r, k, temp;
    int mark;
    if ( s >= e ) return 0;
    res = l = r = 0;
    mark = 1;
    if ( st[s] == 'M' ) {
        k = 0;
        for ( i = s; i < e && mark; ++i ) {
            if ( st[i] == '(' ) ++k;
            else if ( k == 1 && st[i] == ',' ) {
                t1 = dfs(s+4, i, l);
                j = i;
                for ( i; i < e; ++i ) {
                    if ( st[i] == '(' ) ++k;
                    else if ( st[i] == ')' ) --k;
                    if ( k == 0 ) {
                        t2 = dfs(j+1, i, r);
                        if ( t1 > t2 ) {
                            res = t1;
                            num += l*2 + r;
                        } else {
                            res = t2;
                            num += r*2 + l;
                        }
                        mark = 0;
                        break;
                    }
                }
            }
            else if ( st[i] == ')' ) --k;
        }
    }
    temp = 0;
    for ( ; st[i] != 'M' && i < e; ++i ) {
        if ( st[i] == '+' ) {
            ++num;
            res += temp;
            temp = 0;
            continue;
        }
        else if ( st[i] >= '0' && st[i] <= '9' ) temp = temp*10+st[i]-'0';
    }
    res += temp;
    if ( i < e ) return res + dfs(i,e, num);
    return res;
}

int main() {
    int T, res, t;
    scanf("%d%*c", &T);
    while ( T-- ) {
        gets(st);
        res = 0;
        t = dfs(0, strlen(st), res);
        printf("%d %d\n", t, res);
    }
    return 0;
}