C++博客-广交天下ACM爱好者-文章分类-Timus题解

Timus 1087(recursion game)

tw — Tue, 18 Jan 2011 07:23:00 GMT

1087. The Time to Take Stones

Time Limit: 1.0 second
Memory Limit: 16 MB

You probably know the game where two players in turns take 1 to 3 stones from a pile. Looses the one who takes the last stone. We'll generalize this well known game. Assume that both of the players can take not 1, 2 or 3 stones, but k₁, k₂, …, k_m ones. Again we'll be interested in one question: who wins in the perfect game. It is guaranteed that it is possible to make next move irrespective to already made moves.

Input

The first line contains two integers: n and m (1 ≤ n ≤ 10000; 1 ≤ m ≤ 50) — they are an initial amount of stones in the pile and an amount of numbers k₁, …, k_m. The second line consists of the numbers k₁, …, k_m, separated with a space (1 ≤ k_i ≤ n).

Output

Output 1, if the first player (the first to take stones) wins in a perfect game. Otherwise, output 2.

Sample

input	output
17 3 1 3 4	2

Problem Author: Anton Botov
Problem Source: The 3rd high school children programming contest, USU, Yekaterinburg, Russia, March 4, 2001

/*
"Looses the one who takes the last stone" —— the one takes the last
stone loses the game !
*/
#include <stdio.h>
#include <memory>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>
#include <set>
#include <queue>
#include <time.h>
#include <limits>
using namespace std;
#define N 10005
#define M 55
int step[M], n, m;
bool state[N];
bool input(){
    if(scanf("%d%d", &n, &m) == EOF) return false;
    int i;
    for(i = 0; i < m; i++) scanf("%d", step+i);
    return true;
}
void solve(){
    memset(state, 0, sizeof(bool) * (n+1));
    int i, j, u;
    for(i = 1; i <= n; i++){
        for(j = 0; j < m; j++){
            if((u = i - step[j]) >= 0){
                if(u == 0){
                    state[i] = false;
                }else{
                    if(!state[u]){
                        state[i] = true;
                        break;
                    }
                }
            }
        }
    }
    //for(i = 0; i <= n; i++) printf("s[%d]=%d\n", i, state[i]);
    if(state[n]) printf("1\n");
    else printf("2\n");
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif
    while(input()) solve();
    return 0;
}

tw 2011-01-18 15:23 发表评论

Timus 1045 (game on a tree)

tw — Mon, 17 Jan 2011 06:13:00 GMT

#include <stdio.h>
#include <memory>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>
#include <set>
#include <queue>
#include <time.h>
#include <limits>
using namespace std;
#define N 1005
struct e{
    int v;
    e* nxt;
}es[N*2];
e* fir[N];
int en, n, st, l;
bool state[N];
inline void add_e(int u, int v){
    es[en].v = v; es[en].nxt = fir[u]; fir[u] = &es[en++];
}
bool input(){
    if(scanf("%d%d", &n, &st) == EOF) return false;
    st--;
    int i, u, v;
    for(i = 0; i < n; i++) fir[i] = NULL;
    en = 0;
    for(i = 1; i < n; i++){
        scanf("%d%d", &u, &v);
        u--;
        v--;
        add_e(u, v);
        add_e(v, u);
    }
    return true;
}
void dfs(int u, int fa){
    e* cur;
    int v;
    bool hasSon = false;
    for(cur = fir[u]; cur; cur = cur->nxt){
        if((v = cur->v) != fa){
            hasSon = true;
            dfs(v, u);
        }
    }
    if(!hasSon){
        state[u] = false;   //必败态
    }else{
        state[u] = false;
        for(cur = fir[u]; cur; cur = cur->nxt){
            if((v = cur->v) != fa){
                if(!state[v]){
                    if(u == st){
                        if(state[u]){
                            if(l > v + 1) l = v + 1;
                        }else{
                            state[u] = true;
                            l = v + 1;
                        }
                    }else state[u] = true;
                }
            }
        }
    }
}
void solve(){
    dfs(st, -1);
    if(state[st]) printf("First player wins flying to airport %d\n", l);
    else printf("First player loses\n");
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif
    while(input()) solve();
    return 0;
}

tw 2011-01-17 14:13 发表评论

Timus 1023(Bash Game)

tw — Mon, 17 Jan 2011 05:24:00 GMT

// bash game
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int cal(int a){
int b, ans;
ans = inf;
for(b = 1; b * b <= a; b++){
  if(!(a % b)){
   if(b > 2 && b < ans) ans = b;
   if((a / b) > 2 && (a / b) < ans) ans = (a / b);
  }
}
if(ans < inf) return ans - 1;
else return a - 1;
}
int main(){
#ifndef ONLINE_JUDGE
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
init();
int a;
while(~scanf("%d", &a)){
  printf("%d\n", cal(a));
}
return 0;
}

tw 2011-01-17 13:24 发表评论