A、夹娃娃

签到题，前缀和的模板题，没什么坑点，极限都不炸

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) vv.begin(), vv.end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
ll a[N];

int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i) {
        a[i] = read();    a[i] += a[i - 1];
    }
    while (m--) {
        int l = read(), r = read();
        write(a[r] - a[l - 1]), putchar(10);
    }
    return 0;
}

B、莫的难题

进制转换，稍微列举一下前面的一些数你就会发现，第1位的是之下,第2位的之下，第3位的之下，把熟悉的10进制改成五进制了！不过有个坑点五进制开始是1，而不是0，记得计算进制减掉一个1，最后放进一个栈里面去输出即可，非常优秀的题目阿。就是题目表示的有点不明确，求第小的数。线性求下阶乘和逆元就行了。

1 2 3 5 9  //5个数
11 12 13 15 19
21 22 23 25 29
31 32 33 35 39
...
91 92 93 95 99  //5*5+5个数
111 ...
                // 5^3+5^2+5个数
#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) vv.begin(), vv.end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 100 + 7;
string s = "12359";
ll jc[N], inv[N];
int ans[N];

ll C(int n, int m) {
    return jc[n] * inv[n - m] % MOD * inv[m] % MOD;
}

int main() {
    jc[0] = 1;
    for (int i = 1; i < N; ++i)
        jc[i] = jc[i - 1] * i % MOD;
    inv[N - 1] = qpow(jc[N - 1], MOD - 2, MOD);
    for (int i = N - 2; ~i; --i)
        inv[i] = inv[i + 1] * (i + 1) % MOD;
    int T = read();
    while (T--) {
        int n = read(), m = read();
        int c = C(n, m);
        stack<int> st;
        while (c > 0) {
            --c;
            st.push(c % 5);
            c /= 5;
        }
        while (st.size())
            printf("%c", s[st.top()]), st.pop();
        putchar(10);
    }
    return 0;
}

C、不平衡数组

我写的code有点问题，先放着，等我去康康正解再码正确的代码把。
待补

D、数列统计

写个暴力dfs应该很好写吧，IOI先拿10分，再看看能不能找到规律）没办法菜鸡数论不厉害，只能这样解题了

ll ans;
int n, l;
void dfs(int maxi, int len) {
    if (!len) {
        ++ans;
        return;
    }
    for (int i = maxi; i; --i)
        dfs(i, len - 1);
}

int main() {
    int T = read();
    while (T--) {
        n = read(), l = read();
        ans = 0;
        dfs(n, l - 1);
        write(ans), putchar(10);
    }
    return 0;
}

/*
1 1:     1
1 2:     1
1 3:     1
1 4:     1
2 2:     2
2 3:     3
2 4:     4
3 3:     6
3 4:     10
4 4:     20
10 10:     48620
*/

打表出来，你会发现，n，l交换顺序答案是不变的，还有一点是n，l相等的时候是中心二项式的系数。那居然看到了二项式肯定往组合数猜，而且题目给的所以计算必须快，dp肯定是走不通的，这时候一位队友奇迹般地推出的公式救了我们，直接得到答案就是,zqf大佬天下无双！！
图片说明 

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) vv.begin(), vv.end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 911451407;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');  int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';     tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;   while (b) { if (b & 1)  ans *= a;       b >>= 1;      a *= a; }   return ans; }   ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 2e6 + 7;
ll jc[N], inv[N];

ll C(int n, int m) {
    return jc[n] * inv[n - m] % MOD * inv[m] % MOD;
}

int main() {
    jc[0] = 1;
    for (int i = 1; i < N; ++i)
        jc[i] = jc[i - 1] * i % MOD;
    inv[N - 1] = qpow(jc[N - 1], MOD - 2, MOD);
    for (int i = N - 2; i; --i)
        inv[i] = inv[i + 1] * (i + 1) % MOD;
    int t = read();
    while (t--) {
        ll l = read(), x = read();
        ll ans = jc[x + l - 2] * inv[x - 1] % MOD * inv[l - 1] % MOD;
        write(ans), putchar(10);
    }
}