寒假训练1解题报告

1.CodeForces 92A

给一堆围成圈的小朋友发饼干，小朋友为1~n号，第几号小朋友每次拿多少块饼干，问最后剩多少饼干

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
   // freopen("a.in" , "r" , stdin);
    int n , m;
    while(scanf("%d%d" , &n , &m) == 2)
    {
        int tmp = (1 + n ) * n / 2;
        m %= tmp;
        for(int i=1 ; i<=n ; i++){
            if(m >= i) m-=i;
            else break;
        }
        printf("%d\n" , m);
    }
    return 0;
}

2.CodeForces 96A

用0，1分别代表己方和对方球员，如果有7个及以上的同队队员站一起会危险，问是否处于危险状态

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 105;
char s[N];
int pos[N];

int main()
{
   // freopen("a.in" , "r" , stdin);
    while(scanf("%s" , s) != EOF)
    {
        int len = strlen(s);
        for(int i=0 ; i<len ; i++){
            if(s[i] == '0') pos[i] = 0;
            else pos[i] = 1;
        }

        int flag = pos[0] , cnt = 1;
        bool ok = true;
        for(int i=1 ; i<len ; i++){
            if(pos[i] == flag){
                cnt++;
                if(cnt >= 7){
                    ok = false;
                    break;
                }
            }
            else{
                flag = pos[i];
                cnt = 1;
            }
        }

        if(ok) puts("NO");
        else puts("YES");
    }
    return 0;
}

3.CodeForces 71A

将字符串按某种规则缩写

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 105;
char s[N];

int main()
{
  //  freopen("a.in" , "r" , stdin);
    int T;
    scanf("%d" , &T);
    while(T--)
    {
        scanf("%s" , s);
        int len = strlen(s);
        if(len <= 10){
            printf("%s\n" , s);
            continue;
        }
        printf("%c%d%c\n" , s[0] , len-2, s[len-1]);
    }
    return 0;
}

4.CodeForces 71B

枚举所有可能的值就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 105;
int val[N];

int main()
{
   // freopen("a.in" , "r" , stdin);
    int n , k , t;
    while(scanf("%d%d%d" , &n , &k , &t) == 3)
    {
        int tmp = k*n*t , p , q;
        memset(val , 0 , sizeof(val));
        for(p=0 ; p<=n ; p++){
            int flag = 0;
            for(q=0 ; q<=k ; q++){
                int tmp1 = 100*p*k + 100 * q;
              //  cout<<"tmp: "<<tmp<<" "<<tmp1<<endl;
                if(tmp >= tmp1 && tmp<tmp1+100){
                    flag= 1;
                    break;
                }
            }
            if(flag) break;
        }
      //  cout<<p<<" "<<q<<endl;
        for(int i=1 ; i<=p ; i++)
            val[i] = k;
        val[p+1] = q;
        for(int i=1 ; i<=n ; i++)
            if(i == 1) printf("%d" , val[i]);
            else printf(" %d" ,val[i]);
        puts("");
    }
    return 0;
}

5.CodeForces 71C

枚举所有的因子，然后令每个数都对这个因子取模，观察模相同的数是否等于总数除以因子

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 100005;
int vis[N] , mul[N] , cnt , a[N] , tot[N];

void cal(int n)
{
    int t = (int)sqrt(n);
    cnt = 0;
    if(n % 4==0){
        mul[cnt++] = 4;
    }
    while(n % 2 == 0) n>>=1;
    for(int i=3 ; i<=t ; i++){
        if(n % i == 0){
            mul[cnt++] = i;
            while(n % i == 0)n/=i;
        }
        if(i > n) break;
    }
    if(n>1) mul[cnt++] = n;
}

int main()
{
  //  freopen("a.in" , "r" , stdin);
    int n , x;
    while(~scanf("%d" , &n))
    {
        int k=0;
        for(int i=0 ; i< n;i++){
            scanf("%d" , &x);
            if(x == 1) a[k++] = i;
        }
        cal(n);
        int flag = 0;

        for(int i=0 ; i<cnt ; i++)
        {
            int tmp = n/mul[i];
            memset(tot , 0 , sizeof(tot));
            for(int j=0 ; j<k ; j++){
                int pp = a[j] % tmp;
                tot[pp] ++;
                if(tot[pp] == mul[i]){
                    flag = 1;
                    break;
                }
            }
            if(flag) break;
        }
        if(flag) puts("YES");
        else puts("NO");
    }
    return 0;
}

6.CodeForces 71D

纯模拟，代码量略长

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 60;

char str[4];
int num[N][N] , vis[N];
int cnt; //记录可行矩阵的个数

struct Node{
    int x,y;
}node[60];

Node pos1 , pos2;
Node squre1 , squre2;

int change1(char c)
{
    if(c == 'C') return 0;
    else if(c == 'D') return 1;
    else if(c == 'H') return 2;
    else if(c == 'S') return 3;
}

int change2(char c)
{
    if(c == 'A') return 0;
    else if(c == 'T') return 9;
    else if(c == 'J') return 10;
    else if(c == 'Q') return 11;
    else if(c == 'K') return 12;
    else {
        return (int)(c-'1');
    }
}

bool ok(int i , int j)
{
    int flag1 = 1 , flag2 = 1;
    int mod[20];
    memset(mod, 0 , sizeof(mod));
    for(int p=0 ; p<3 ; p++){
        for(int q=0 ; q<3 ; q++){
            int t = num[i+p][j+q] % 13;
            if(mod[t]){
                flag1 = 0;
                break;
            }
            mod[t] = 1;
        }
        if(!flag1) break;
    }
    if(flag1) return true;

    int col = num[i][j] / 13;
    for(int p=0 ; p<3 ; p++){
        for(int q=0 ; q<3 ; q++){
            int t = num[i+p][j+q] / 13;
            if(t != col){
                flag2 = 0;
                break;
            }
        }
        if(!flag2) break;
    }

    if(flag2) return true;
    else return false;
}

bool ok1(Node m1 , Node m2)
{
    if(m1.x > m2.x+2) return true;
    if(m1.x+2 < m2.x) return true;
    if(m1.y+2 < m2.y) return true;
    if(m1.y > m2.y+2) return true;
    return false;
}

void get_martrix(int n , int m)
{
    cnt = 0;
    for(int i=1 ; i<=n-2 ; i++){
        for(int j=1 ; j<=m-2 ; j++){
            if(ok(i,j)){
                node[cnt].x = i;
                node[cnt++].y = j;
            }
        }
    }
}

char* intTostring(int x)
{
    char *s = new char [3];
    int t1 = x/13;
    if(t1 == 0) s[1] = 'C';
    else if(t1 == 1) s[1] = 'D';
    else if(t1 == 2) s[1] = 'H';
    else if(t1 == 3) s[1] = 'S';

    int t2 = x%13;
    if(t2 == 0) s[0] = 'A';
    else if(t2 == 9) s[0] = 'T';
    else if(t2 == 10) s[0] = 'J';
    else if(t2 == 11) s[0] = 'Q';
    else if(t2 == 12) s[0] = 'K';
    else {
        s[0] = (char)(t2+'1');
    }
    s[2] = '\0';
    return s;
}

int main()
{
   // freopen("a.in" , "r" , stdin);
    int n , m;
    while(scanf("%d%d" , &n , &m) == 2)
    {
        bool flag1 = false , flag2 = false;
        int ans1 , ans2 , card1 , card2;//记录最后选中的在第几号节点
        pos1.x = pos1.y = pos2.x = pos2.y = 0;
        memset(vis , 0 , sizeof(vis));

        for(int i=1 ; i<=n ; i++)
            for(int j = 1 ; j<=m ; j++)
            {
                scanf("%s" , str);
                if(str[1] == '1'){
                    pos1.x = i;
                    pos1.y = j;
                    flag1 = true;
                }
                else if(str[1] == '2'){
                    pos2.x = i;
                    pos2.y = j;
                    flag2 = true;
                }
                else{
                    num[i][j] = 13*change1(str[1]) + change2(str[0]);
                    vis[num[i][j]] = 1;
                }
            }
       /* for(int i=1 ; i<=n ; i++){
            for(int j = 1 ; j<=m ; j++)
                cout<<num[i][j]<<" ";

            puts("");
        }*/
        bool find = false;
        int time=0;
        if(flag1) time++;
        if(flag2) time++;
        for(int p=0 ; p<52 ; p++){
            for(int q=0 ; q<52 ; q++){
                if(p == q) continue;
                int change = time;
                if(flag1 && !vis[p]) num[pos1.x][pos1.y] = p,change--;
                if(flag2 && !vis[q]) num[pos2.x][pos2.y] = q,change--;

                if(!change)
                {
                    get_martrix(n , m);

                    for(int i=0 ; i<cnt ; i++){
                        for(int j=i+1 ; j<cnt ; j++){
                            if(ok1(node[i] , node[j])){
                                find=true;
                                ans1 = i , ans2 = j;
                                card1 = p , card2 = q;
                                break;
                            }
                        }
                        if(find) break;
                    }

                }
                if(find) break;
            }
            if(find) break;
        }

        if(!find)
            puts("No solution.");
        else{
            puts("Solution exists.");
            if(!flag1 && !flag2){
                puts("There are no jokers.");
            }
            else if(flag1 && flag2){
                char *s1 = new char [3];
                char *s2 = new char [3];
                s1 = intTostring(card1);
                s2 = intTostring(card2);
                printf("Replace J1 with %s and J2 with %s.\n" , s1 , s2);
            }
            else if(flag1){
                char *s1 = new char [3];
                s1 = intTostring(card1);
                printf("Replace J1 with %s.\n" , s1);
            }
            else if(flag2){
                char *s1 = new char [3];
                s1 = intTostring(card2);
                printf("Replace J2 with %s.\n" , s1);
            }
            printf("Put the first square to (%d, %d).\n" , node[ans1].x , node[ans1].y);
            printf("Put the second square to (%d, %d).\n" , node[ans2].x , node[ans2].y);
        }
    }
    return 0;
}

7.CodeForces 88E

一道简单的nim博弈，sg函数值的计算过程中注意保存得到的最小堆数，过程可以用异或的前缀和帮助自己提高效率

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace std;
const int N = 100010;

int sg[N] , sum[N] , minn[N] , vis[N];

void init_sg()
{
    sg[0] = sg[1] = sg[2] = 0;
    sum[0] = sum[1] = sum[2] = 0;
    memset(minn , 0x3f , sizeof(minn));
    for(int n=3 ; n<=100000 ; n++){
        int t = 2*n;
        int factor = (int)(sqrt(t+0.5));
        for(int k=2 ; k<=factor ; k++){
            if(t % k == 0 && t/k+1-k>0 && !((t/k+1-k)&1)){
                int a = (t/k+1-k)/2;
                int p = sum[a+k-1]^sum[a-1];
                vis[p] = n;
                if(p == 0) minn[n] = min(minn[n] , k);
            }
        }
        //找没有出现过的最小的sg函数
        int v = 0;
        while(1){
            if(vis[v] != n){
                sg[n] = v;
                break;
            }
            v++;
        }
        //记录当前的sum[]前缀
        sum[n] = sum[n-1]^sg[n];
    }
}

int main()
{
  //  freopen("a.in" , "r" , stdin);
    init_sg();
    int n;
    while(scanf("%d" , &n) == 1)
    {
        if(!sg[n]){
            puts("-1");
            continue;
        }
        printf("%d\n" , minn[n]);
    }
    return 0;
}