题一：A Good Game

题解：利用前缀和，求出所有区间值，然后操作排序，贪心求解即可
知识点：前缀和 + 贪心

AC代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<string.h>
#include<string>
using namespace std;
const int N = 1e6 + 15;
#define ll long long
#define inf 0x3f3f3f3f
ll a[N], sum[N];
ll b[N];
int main()
{
   
	int t; cin >> t;
	while (t--) {
   
		int n, m;
		cin >> n >> m;
		for (int i = 0; i < n; i++)
			cin >> a[i];
		sum[0] = a[0];
		for (int i = 1; i < n; i++)
			sum[i] = sum[i - 1] + a[i];
		int k = 0;
		for (int i = 0; i < m; i++) {
   
			int l, r;
			cin >> l >> r;
			int x = sum[r - 1] - sum[l - 2];
			b[k++] = x;
		}
		sort(b, b + k);
		ll res = 0;
		for (int i = 1; i <= k; i++)
			res += i * b[i - 1];
		cout << res << endl;
	}
	return 0;
}

题二：Prefix

题解：map瞎搞，记录前缀的数量，暴力就可以过，对于我这种只会做签到题的菜鸡。不过这题更好的解法是字典树。
注意：记得把所有的有关数据都开ll，以免相乘的时候出现精度丢失

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<string.h>
#include<string>
using namespace std;
const int N = 5e5 + 15;
#define ll long long
#define inf 0x3f3f3f3f
ll val[N];
string str[N];
map<string, ll>mp;
int main()
{
   
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);
	int n, m; cin >> n >> m;
	ll s_val[30];
	for (int i = 0; i < 26; i++)
		cin >> s_val[i];
	for (int i = 0; i < n; i++) {
   
		cin >> str[i];
		int str_len = str[i].length();
		string s;
		val[i] = 1;
		for (int j = 0; j < str_len; j++) {
   
			s += str[i][j];
			mp[s] += 1;
			val[i] = (val[i] * s_val[str[i][j] - 'a']) % m;
		}
	}
	for (int i = 0; i < n; i++) {
   
		string s;
		int v = 1;
		int ans = 0;
		for (int j = 0; j < str[i].length(); j++) {
   
			v = (v * s_val[str[i][j] - 'a']) % m;
			s += str[i][j];
			if (v > val[i])
				ans += mp[s];
		}
		cout << ans << " ";
	}
	cout << endl;
	return 0;
}

题三：Sequence

题解：线段树的更新，改点，异或的应用

AC代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<string.h>
#include<string>
using namespace std;
const int N = 2e5 + 15;
#define ll long long
#define inf 0x3f3f3f3f
struct Tree {
   
	int l, r;
	int val,sum;
}tree[N * 2];
int a[N];
void pushup(int cur) {
   
	if ((tree[cur * 2 + 1].l - tree[cur].l) % 2 == 0)
		tree[cur].val = tree[cur * 2].val ^ tree[cur * 2 + 1].val;
	else
		tree[cur].val = tree[cur * 2].val ^ (tree[cur * 2 + 1].sum ^ tree[cur * 2 + 1].val);
	tree[cur].sum = tree[cur * 2].sum ^ tree[cur * 2 + 1].sum;
}
void build(int l, int r, int step) {
   
	int mid = (l + r) / 2;
	tree[step].l = l, tree[step].r = r;
	if (l == r) {
   
		tree[step].val = a[l];
		tree[step].sum = a[l];
		return;
	}
	build(l, mid, step * 2);
	build(mid + 1, r, step * 2 + 1);
	pushup(step);
}
void query(int l, int r, int &x, int &y, int step) {
   
	if (tree[step].l == l && tree[step].r == r) {
   
		x = tree[step].val, y = tree[step].sum;
		return;
	}
	int mid = (tree[step].l + tree[step].r) / 2;
	if (r <= mid)
		query(l, r, x, y, step * 2);
	else if (l > mid)
		query(l, r, x, y, step * 2 + 1);
	else {
   
		int a, b, c, d;
		query(l, mid, a, b, step * 2);
		query(mid + 1, r, c, d, step * 2 + 1);
		if ((tree[step * 2 + 1].l - l) % 2 == 0) {
   
			x = a ^ c;
			y = b ^ d;
		}
		else {
   
			x = a ^ (d ^ c);
			y = b ^ d;
		}
	}

}
void update(int tar, int val, int step) {
   
	int l = tree[step].l, r = tree[step].r;
	if (l == r) {
   
		tree[step].val = val;
		tree[step].sum = val;
		return;
	}
	int mid = (l + r) / 2;
	if (tar <= mid)
		update(tar, val, step * 2);
	else
		update(tar, val, step * 2 + 1);
	pushup(step);
}
int main()
{
   
	int t; 
	scanf("%d", &t);
	int k = 1;
	while (t--) {
   
		int n, m; 
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		build(1, n, 1);
		printf("Case #%d:\n", k++);
		for (int i = 0; i < m; i++) {
   
			int op, x, y;
			scanf("%d%d%d", &op, &x, &y);
			if (op == 0)
				update(x, y, 1);
			else {
   
				if ((y - x + 1) % 2 != 0) {
   
					int a, b;
					query(x, y, a, b, 1);
					printf("%d\n", a);
				}
				else
					printf("0\n");
			}
		}
	}
	return 0;
}