写在前面&总结：
(LuckyBlock) 良心出题人！暴力分给了 (120pts)
(T1) 貌似是个结论题，最后知道怎么算了，用前缀和搞了两下，写挂了就很草，最后只能靠暴力拿了 (30pts)
(T2) 显然数论题，但是我不会化简/kk，不过用前缀和优化了一下暴力 (40pts);
(T3) 依旧暴力，(O(n^2)) 可拿 (40pts)
(T4) 部分分 (5pts) (+) (lps) 随机输出 ('7') (5pts);
不会打暴力的 (OIer) 不是好 (OIer) ，不打暴力的都是**（没错说的就是CSP2020上的我
感觉 (LuckyBlock) 这次考察的芝士点比较详细，先粘个题解网址咕着，以后再补
T1
改自 CF1422C
发现可以当做取出一段序列，把剩下的拼接起来
在拼接过程中发现有许多部分是重复计算的，可以用后缀和和开个 (10^i) 的数组预处理
算了我也说不清楚直接放代码吧嘻嘻
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int MAXN = 2e6+6;
const int INF = 1;
const int mod = 1e9+7;
LL hsum[MAXN], jc[MAXN];
char ch[MAXN];
LL ans;
int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return s * w;
}
LL quick(LL x, LL p, LL mod){
LL res = 1;
for( ; p; p >>= 1){
if(p & 1) res = res * x % mod;
x = x * x % mod;
}
return res;
}
signed main()
{
cin>>ch;
LL len = strlen(ch);
int cnt = 0;
cnt = 1; jc[0] = 1;
for(int i = 1; i <= len; ++i){
jc[i] = jc[i - 1] * 10 % mod;
}
cnt = 0;
for(int i = len - 1; i >= 0; --i){//求后缀数
cnt = (cnt + (ch[i] - '0') * jc[len - 1 - i] % mod) % mod;
hsum[i] = cnt;
// cout<<hsum[i]<<' ';
}
LL val = 0, sum = 0;
for(int i = 0; i < len; ++i){
ans += sum * jc[len - 1 - i] % mod;
ans += i * hsum[i] % mod;
ans %= mod;
val = (10 * val % mod + ch[i] - '0') % mod;
sum = (sum + val) % mod;
}
printf("%lld", ans % mod);
return 0;
}
暴力部分核心代码：
for(int l = 0; l < len; ++l){
LL cnt = 0;
for(int r = l; r < len; ++r){
for(int i = 0; i < len; ++i){
if(i < l || i > r){
cnt = (cnt * 10 % mod + ch[i] - '0') % mod;
}
}
ans = (ans + cnt) % mod;
}
}
T2
(40pts) 暴力：
求 (gcd(i, n)) 是有多次求的，可以用前缀和预处理 (2000) 个点然后 (O(n)) 出结果
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1e6+6;;
const int INF = 1;
const int mod = 998244353;
int T, l, r;
int gcd[2010][2010];
int sum[2020];
int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return s * w;
}
int Gcd(int x, int y){
return x % y == 0 ? y : Gcd(y, x % y);
}
void init(){
for(int i = 1; i <= 2000; ++i){
for(int j = 1; j <= i; ++j){
// a[i][j] = Gcd(j, i);
sum[i] = (sum[i] + Gcd(j, i)) % mod ;
}
}
}
int main()
{
init();
T = read();
while(T--){
l = read(), r = read();
int ans = 0;
for(int i = l; i <= r; ++i){
ans = (ans + sum[i]) % mod;
}
printf("%d
", ans);
}
return 0;
}
(100pts) 正解
考虑化一下 (f) 。
[f(n) = sum_{i = 1}^{n} gcd (i, n)
]
考虑对于每一个 (1 sim n) 的值，能作为多少数对的 (gcd) ，于是有：
[f(n) = sum_{i = 1}^{n}dsum_{i = 1}^{n}[gcd(i, n) = d]
]
发现 (gcd(i, n) = d) 的必要条件是 (d mid n) ，原式可以改为：
[f(n) = sum_{d mid n} d sum_{i = 1}^{n}[gcd(i, n) = d]
]
考虑什么样的 (i) ，满足 (gcd(i, n) = d) ，显然当且仅当 (i = kd(k in mathbb{N^*})) ，且 (gcd(k, frac{n}{d}) = 1) 是满足条件。为保证 (i le n) ，有 (k le leftlfloorfrac{n}{d} ight floor) 。
于是考虑把 (d) 提出来，改为枚举上述的 (k) ，原式等于：
[f(n) = sum_{d|n} d sum_{k=1}^{leftlfloorfrac{n}{d}
ight
floor}left[gcdleft(k,frac{n}{d}
ight)=1
ight]
]
考虑后面一个 (sum) 的实际意义，表示 (1 sim frac{n}{d}) 中与 (frac{n}{d}) 互质的个数，符合欧拉函数的定义，于是原式等于：
[f(n) = sum_{d|n} d cdot varphileft(frac{n}{d}
ight)
]
线性筛预处理 (varphi) 后，有埃氏筛即可筛出 (1 sim 10^{6}) 的所有的 (f) 。
做个前缀和即可回答区间询问
复杂度 (O(n log n + m))
（Solution来自LuckyBlock的题解）
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1e6+5;
const int kMax = 1e6;
const int INF = 1;
const int mod = 998244353;
int p_cnt, p[MAXN], phi[MAXN];
int f[MAXN], sum[MAXN];
bool vis[MAXN];
int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return s * w;
}
void init(){
phi[1] = 1;
for(int i = 2; i <= kMax; ++i){
if(!vis[i]){
p[++p_cnt] = i;
phi[i] = i - 1;
}
for(int j = 1; j <= p_cnt && i * p[j] <= kMax; ++j){
vis[i * p[j]] = true;
if(i % p[j] == 0){
phi[i * p[j]] = phi[i] * p[j];
break;
}
phi[i * p[j]] = phi[i] * (p[j] - 1);
}
}
for(int i = 1; i <= kMax; ++i){
for(int j = i; j <= kMax; j += i){
f[j] = (f[j] + 1ll * phi[i] * (j / i) % mod) % mod;
}
}
for(int i = 1; i <= kMax; ++i){
sum[i] = (sum[i - 1] + f[i]) % mod;
}
}
int main()
{
init();
int m = read();
while(m--){
int l = read(), r = read();
printf("%d
", (sum[r] - sum[l - 1] + mod) % mod);
}
}
T3
(40pts) 暴力：
暴力更改区间起点暴力求所有情况的最大值即可
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1;
const int INF = 1;
const int mod = 1;
int n;
char a[50100], b[50010];
int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return s * w;
}
int main()
{
n = read();
cin>>a>>b;
int ans = -1;
for(int i = 0; i < n; ++i){
int cnt = 0;
for(int j = 0; j < n; ++j){
int x = i + j;
x = (x >= n ? x - n : x);
if(a[j] == '1' && b[x] == '1'){
cnt++;
}
}
ans = max(ans, cnt);
}
printf("%d", ans);
return 0;
}
(100pts) 正解：
利用 bitset 容器，详细介绍请看 Oi-Wiki
/*
Work by: Suzt_ilymics
Knowledge: ??
Time: O(??)
*/
#include<iostream>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<string>
#include<algorithm>
#include<bitset>
using namespace std;
const int MAXN = 1e5+10;
const int INF = 1;
const int mod = 1;
int n, ans;
char s1[MAXN], s2[MAXN];
bitset <MAXN> a, b, c;
int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + ch - '0' , ch = getchar();
return s * w;
}
void max(int &x, int y){if(x < y) x = y; }
int main()
{
n = read();
scanf("%s", s1 + 1);
scanf("%s", s2 + 1);
for(int i = 1; i <= n; ++i){
a[i] = (s1[i] == '1');
b[i] = (s2[i] == '1');
}
for(int i = 1; i <= n; ++i){
b[n + 1] = b[1];
b >>= 1;
max(ans, (a & b).count());
}
printf("%d", ans);
return 0;
}
T4
改自 CF1422C
只拿了部分分就不粘码了嘻嘻