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BestCoder Round #41

編輯：關於C++

題目傳送：BestCoder Round #41

A、ZCC loves straight flush

思路：簡單題，不過剛開始沒看清題，wa了好幾次，然後才發現輸入不連續也可以，就是說每個同一花色的牌都可以放在一塊，不用在意輸入順序，感覺這裡題目應該說清楚點好些

AC代碼（略挫，比賽時寫的都比較亂）：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

char a[5][5];

int fun(int *p, int n) {
    sort(p, p + n);
    int ret = 1;
    if(p[0] != 1) {
        for(int i = 1; i <= 9; i ++) {
            int cur = 0;
            for(int j = 0; j < n; j ++) {
                if(p[j] <= i + 4 && p[j] >= i) {
                    cur ++;
                }
                ret = max(ret, cur);
            }
        }
    }
    else if(p[0] == 1) {
        for(int i = 1; i <= 10; i ++) {
            int cur = 0;
            for(int j = 0; j < n; j ++) {
                if(p[j] <= i + 4 && p[j] >= i) {
                    cur ++;
                }
            }
            if(i == 10) cur ++;
            ret = max(ret, cur);
        }
    }
    return ret;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        scanf("%s %s %s %s %s", a[0], a[1], a[2], a[3], a[4]);
        int cur = 1;
        int num[5];
        memset(num, 0, sizeof(num));
        for(int i = 0; i < 5; i ++) {
            int len = strlen(a[i]);
            for(int j = 1; j < len; j ++) {
                num[i] = num[i] * 10 + a[i][j] - '0';
            }
        }
        /*for(int i = 0; i < 5; i ++) {
            cout << num[i] << " ";
        }
        cout << endl;*/
        int A[5], B[5], C[5], D[5];
        int an = 0, bn = 0, cn = 0, dn = 0;
        for(int i = 0; i < 5; i++) {
            if(a[i][0] == 'A') A[an ++] = num[i];
            else if(a[i][0] == 'B') B[bn ++] = num[i];
            else if(a[i][0] == 'C') C[cn ++] = num[i];
            else if(a[i][0] == 'D') D[dn ++] = num[i];
        }
        
        int ans = 1;
        ans = max(ans, fun(A, an));
        ans = max(ans, fun(B, bn));
        ans = max(ans, fun(C, cn));
        ans = max(ans, fun(D, dn));
        
        printf("%d\n", 5 - ans);
    }
    return 0;
}

B、ZCC loves strings

思路：找規律，簡單博弈，但是坑點較多，總共有Cn取2中情況，然後Miss G贏得情況有奇數和偶數湊一塊的時候，還有選擇相同字符串的時候，然後特判一下ans是否為0，再取個gcd，記得用longlong，會爆int

AC代碼：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int a[200005];

LL gcd(LL a, LL b) {
    return b == 0 ? a : gcd(b, a % b);
}

int main() {
    ios::sync_with_stdio(false);  
    int T, n;
    scanf("%d", &T);
    while(T --) {
        scanf("%d", &n);
        string tmp;
        LL ji = 0, ou = 0;
        map mmp;
        LL jia = 0;
        for(int i = 0; i < n; i ++) {
            cin >> tmp;
            if(mmp.find(tmp) != mmp.end()) {
                jia += mmp[tmp];
                mmp[tmp] ++;
            }
            else mmp[tmp] = 1;
            
            a[i] = tmp.length();
            if(a[i] & 1) ji ++;
            else ou ++;
        }
        LL ans = ji * ou;
        LL sum = n * (n - 1) / 2;
        
        ans += jia;
        if(ans == 0) {
            printf("0/1\n");
            continue;
        }
        
        LL gg = gcd(ans, sum);
        ans /= gg;
        sum /= gg;
        cout << ans << '/' << sum << endl;
    }
    return 0;
}