#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
#include<queue>
#include<vector>
using namespace std;
#define ll long long int
#define INF 0x3f3f3f3f
const int maxn = 2e5 + 10;
int f[maxn];
int sum[maxn];
int
 
 find(int x) {
    if (x != f[x]) {
        int tmp = f[x];
        f[x] = find(f[x]);
        sum[x] = (sum[x] ^ sum[tmp]);
    }
    return f[x];
}
int main()
{
    int n, m;
    while (~scanf("%d %d", &n, &m)) {
        for (int i = 0; i <= n; i++) f[i] = i, sum[i] = 0;
        int ans = 0
 
; bool flag = 0;
        for (int i = 1; i <= m; i++) {
            int l, r, s;
            scanf("%d %d %d", &l, &r, &s);
            l--;
            int ra = find(l); int rb = find(r);
            if (ra == rb) {
                if ((sum[l] ^ sum[r]) != s) {
                    flag = 1
 
;
                    printf("%d\n", i);
                }
            }
            else {
                f[ra] = rb;
                sum[ra] = (sum[r] ^ sum[l] ^ s);
            }
        }
        if (!flag)
            printf("-1\n");
    }
    return 0;
}

B Trees on Sale
1.題目的要求可以簡化成，任意多個人要同樣數量的樹苗，都可以滿足。
2.通過打表可以發現捆的數量存在規律！

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

#define  LL long long

int main()
{
    LL n, k;
    while (~scanf("%lld %lld", &n, &k)) {
        LL now = 1;
        LL sum = 0;
        LL aa = 0;
        while (now <= k) {
            LL xian = sum / now;
            LL ans = n - xian;
            sum += ans * now;

            if (ans != 0) {
                aa += ans;
            }
            now = sum / n + 1;
        }
        printf("%lld\n", aa);
    }



    return 0;
}

F Beautiful Land
超大0-1揹包問題，想辦法來模擬0-1揹包的過程，注意中間狀態需要不斷更新，不可保留！記憶體超限！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<string>
#include<algorithm>
#include<queue>

using namespace std;
map<int, long long> mp;
int w;
int n;
long long sum;
typedef struct
{
    int w, v;
}Value;
Value value[150];
priority_queue<int> p;
bool compare(Value v1, Value v2)
{
    return v1.w > v2.w;
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d %d", &n, &w);

        for (int i = 1; i <= n; i++)
        {
            scanf("%d %d", &value[i].w, &value[i].v);
        }
        mp[0] = 0;
        sum = 0;
        sort(value + 1, value + 1 + n, compare);
        for (int i = 1; i <= n; i++)
        {
            if (value[i].v == 0 || value[i].w > w) continue;
            long long maxx = 0;
            for (map<int, long long>::iterator it = mp.begin(); it != mp.end(); ++it)
            {
                int d = it->first;
                if ((long long)d + (long long)value[i].w <= (long long)w && maxx <= it->second + value[i].v) { p.push(d + value[i].w); }
                maxx = max(maxx, it->second + value[i].v);
            }
            while (!p.empty())
            {
                int d = p.top(); p.pop();
                mp[d] = max(mp[d], mp[d - value[i].w] + value[i].v);
                if (sum < mp[d]) { sum = mp[d]; }
            }
        }
        mp.clear();
        printf("%lld\n", sum);

    }
    return 0;
}

G Cards Game
博弈問題：
有N堆牌，每張牌都有自己的權值，A每次可以拿任意一堆的堆頂，B每次可以拿任意一堆的堆底。如果我在某一堆的半堆中有很好的牌，我必不可能被其拿去，若有不好的牌也必不可能被對方拿去，反之亦然。那麼決策的關鍵就來了，對於偶數堆，每個人拿到的牌就可以固定了，但是對於奇數堆的中間元素就有了決策的可能。也就意味著，A和B從大到小排著拿了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define  LL long long
int a[105][105];
int c[105];
int x[105];
bool cmp(int a, int b) {
    return a > b;
}
int main()
{
    int n; int m; int tmp1 = 0, tmp2 = 0;
    scanf("%d", &n); int cnt = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &m);
        for (int i = 1; i <= m; i++) {
            scanf("%d", &x[i]);
        }
        if (m % 2) {
            for (int i = 1; i <= m / 2; i++) {
                tmp1 += x[i];
            }
            for (int i = (m + 1) / 2 + 1; i <= m; i++){
                tmp2 += x[i];
            }
            c[++cnt] = x[(m + 1) / 2];
        }
        else {
            for (int i = 1; i <= m / 2; i++) {
                tmp1 += x[i];
            }
            for (int i = m / 2 + 1; i <= m; i++) {
                tmp2 += x[i];
            }
        }
    }
    sort(c + 1, c + 1 + cnt, cmp);
    for (int i = 1; i <= n; i++) {
        if (i % 2) tmp1 += c[i];
        else tmp2 += c[i];
    }
    printf("%d %d\n", tmp1, tmp2);
    return 0;
}