“科林明伦杯”哈尔滨理工大学第十届程序设计竞赛(同步赛)
B.减成一
思路：
差分的思想，当当前项小于前一项的时候说明这一项可以和前一项同时减去。
代码：
//#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#define INF 0x3f3f3f3f
#define rep(i, a, n) for (int i=(a); i<(n); i++)
#define per(i, a, n) for (int i=(a); i>(n); i--)
typedef long long ll;
const int maxn = 2e5+5;
const int mod = 1e9+7;
using namespace std;
ll a[maxn];
void slove() {
ll n,ans=0; cin >> n;
rep(i,1,n+1) cin >> a[i];
ans = a[1]-1;
rep(i,2,n+1) {
if(a[i] > a[i-1]) ans+=a[i]-a[i-1];
}
cout << ans << '\n';
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
int t; cin >> t; while(t--)
slove();
return 0;
} C.面积
思路：
签到题，按照题意做就好
代码：
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define rep(i, a, n) for (int i=(a); i<(n); i++)
#define per(i, a, n) for (int i=(a); i>(n); i--)
typedef long long ll;
const int maxn = 2e5+5;
const int mod = 1e9+7;
using namespace std;
#define PI 3.14
void slove() {
double x;
cin >> x;
cout << setprecision(2) << (x/2)*(x/2)*2*PI+x*x << '\n';
}
int main()
{
cout.setf(ios::fixed);
ios_base::sync_with_stdio(false); cin.tie(0);
int t; cin >> t; while(t--)
slove();
return 0;
} E.赛马
思路：
先对双方的马排序，根据对手的马，每次从小明的马中找到一个刚好大于对手的马，记入答案。
代码：
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define rep(i, a, n) for (int i=(a); i<(n); i++)
#define per(i, a, n) for (int i=(a); i>(n); i--)
typedef long long ll;
const int maxn = 2e3+5;
const int mod = 1e9+7;
using namespace std;
ll a[maxn], b[maxn];
int f[maxn];
void slove() {
ll n,ans=0; cin >> n;
rep(i, 1, n+1) cin >> a[i];
rep(i, 1, n+1) cin >> b[i];
sort(a+1,a+n+1);
sort(b+1,b+n+1);
rep(i,1,n+1) {
ll p = upper_bound(a+1,a+n+1,b[i])-a;
if (a[p]) {
ans++,a[p]=0;
sort(a+1,a+n+1);
}
}
cout << ans << '\n';
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
int t; cin >> t; while(t--)
slove();
return 0;
} F.三角形
思路：
可以发现不能构成三角形的最短的三条边的长度是1、1、2，第四条边的长度必须大于第二、三项的和，于是可以构造出斐波那契数列，计算前缀和，可以发现题目的数据最大不超过91项
代码：
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define rep(i, a, n) for (int i=(a); i<(n); i++)
#define per(i, a, n) for (int i=(a); i>(n); i--)
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 2e5+5;
const int mod = 1e9+7;
using namespace std;
ull a[maxn], d[maxn];
void slove() {
ull n; cin >> n;
int pos = upper_bound(d,d+92,n)-d;
if (pos == 92) pos--;
cout << pos << '\n';
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
a[0]=a[1]=1;
d[0]=1;
rep(i,2,93) a[i]=a[i-1]+a[i-2];
rep(i,1,93) d[i]=a[i]+d[i-1]; //cout << d[i] << '\n';
int t; cin >> t; while(t--)
slove();
return 0;
} H.直线
思路：
签到题，大数乘法。
代码：
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define rep(i, a, n) for (int i=(a); i<(n); i++)
#define per(i, a, n) for (int i=(a); i>(n); i--)
typedef long long ll;
const int maxn = 2e5+5;
const int mod = 1e9+7;
using namespace std;
const int base = 1000000000;
const int base_digits = 9;
struct bigint {
vector<int> a;
int sign;
/*<arpa>*/
int size(){
if(a.empty())return 0;
int ans=(a.size()-1)*base_digits;
int ca=a.back();
while(ca)
ans++,ca/=10;
return ans;
}
bigint operator ^(const bigint &v){
bigint ans=1,a=*this,b=v;
while(!b.isZero()){
if(b%2)
ans*=a;
a*=a,b/=2;
}
return ans;
}
string to_string(){
stringstream ss;
ss << *this;
string s;
ss >> s;
return s;
}
int sumof(){
string s = to_string();
int ans = 0;
for(auto c : s) ans += c - '0';
return ans;
}
/*</arpa>*/
bigint() :
sign(1) {
}
bigint(long long v) {
*this = v;
}
bigint(const string &s) {
read(s);
}
void operator=(const bigint &v) {
sign = v.sign;
a = v.a;
}
void operator=(long long v) {
sign = 1;
a.clear();
if (v < 0)
sign = -1, v = -v;
for (; v > 0; v = v / base)
a.push_back(v % base);
}
bigint operator+(const bigint &v) const {
if (sign == v.sign) {
bigint res = v;
for (int i = 0, carry = 0; i < (int) max(a.size(), v.a.size()) || carry; ++i) {
if (i == (int) res.a.size())
res.a.push_back(0);
res.a[i] += carry + (i < (int) a.size() ? a[i] : 0);
carry = res.a[i] >= base;
if (carry)
res.a[i] -= base;
}
return res;
}
return *this - (-v);
}
bigint operator-(const bigint &v) const {
if (sign == v.sign) {
if (abs() >= v.abs()) {
bigint res = *this;
for (int i = 0, carry = 0; i < (int) v.a.size() || carry; ++i) {
res.a[i] -= carry + (i < (int) v.a.size() ? v.a[i] : 0);
carry = res.a[i] < 0;
if (carry)
res.a[i] += base;
}
res.trim();
return res;
}
return -(v - *this);
}
return *this + (-v);
}
void operator*=(int v) {
if (v < 0)
sign = -sign, v = -v;
for (int i = 0, carry = 0; i < (int) a.size() || carry; ++i) {
if (i == (int) a.size())
a.push_back(0);
long long cur = a[i] * (long long) v + carry;
carry = (int) (cur / base);
a[i] = (int) (cur % base);
//asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));
}
trim();
}
bigint operator*(int v) const {
bigint res = *this;
res *= v;
return res;
}
void operator*=(long long v) {
if (v < 0)
sign = -sign, v = -v;
if(v > base){
*this = *this * (v / base) * base + *this * (v % base);
return ;
}
for (int i = 0, carry = 0; i < (int) a.size() || carry; ++i) {
if (i == (int) a.size())
a.push_back(0);
long long cur = a[i] * (long long) v + carry;
carry = (int) (cur / base);
a[i] = (int) (cur % base);
//asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));
}
trim();
}
bigint operator*(long long v) const {
bigint res = *this;
res *= v;
return res;
}
friend pair<bigint, bigint> divmod(const bigint &a1, const bigint &b1) {
int norm = base / (b1.a.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.a.resize(a.a.size());
for (int i = a.a.size() - 1; i >= 0; i--) {
r *= base;
r += a.a[i];
int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
int d = ((long long) base * s1 + s2) / b.a.back();
r -= b * d;
while (r < 0)
r += b, --d;
q.a[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return make_pair(q, r / norm);
}
bigint operator/(const bigint &v) const {
return divmod(*this, v).first;
}
bigint operator%(const bigint &v) const {
return divmod(*this, v).second;
}
void operator/=(int v) {
if (v < 0)
sign = -sign, v = -v;
for (int i = (int) a.size() - 1, rem = 0; i >= 0; --i) {
long long cur = a[i] + rem * (long long) base;
a[i] = (int) (cur / v);
rem = (int) (cur % v);
}
trim();
}
bigint operator/(int v) const {
bigint res = *this;
res /= v;
return res;
}
int operator%(int v) const {
if (v < 0)
v = -v;
int m = 0;
for (int i = a.size() - 1; i >= 0; --i)
m = (a[i] + m * (long long) base) % v;
return m * sign;
}
void operator+=(const bigint &v) {
*this = *this + v;
}
void operator-=(const bigint &v) {
*this = *this - v;
}
void operator*=(const bigint &v) {
*this = *this * v;
}
void operator/=(const bigint &v) {
*this = *this / v;
}
bool operator<(const bigint &v) const {
if (sign != v.sign)
return sign < v.sign;
if (a.size() != v.a.size())
return a.size() * sign < v.a.size() * v.sign;
for (int i = a.size() - 1; i >= 0; i--)
if (a[i] != v.a[i])
return a[i] * sign < v.a[i] * sign;
return false;
}
bool operator>(const bigint &v) const {
return v < *this;
}
bool operator<=(const bigint &v) const {
return !(v < *this);
}
bool operator>=(const bigint &v) const {
return !(*this < v);
}
bool operator==(const bigint &v) const {
return !(*this < v) && !(v < *this);
}
bool operator!=(const bigint &v) const {
return *this < v || v < *this;
}
void trim() {
while (!a.empty() && !a.back())
a.pop_back();
if (a.empty())
sign = 1;
}
bool isZero() const {
return a.empty() || (a.size() == 1 && !a[0]);
}
bigint operator-() const {
bigint res = *this;
res.sign = -sign;
return res;
}
bigint abs() const {
bigint res = *this;
res.sign *= res.sign;
return res;
}
long long longValue() const {
long long res = 0;
for (int i = a.size() - 1; i >= 0; i--)
res = res * base + a[i];
return res * sign;
}
friend bigint gcd(const bigint &a, const bigint &b) {
return b.isZero() ? a : gcd(b, a % b);
}
friend bigint lcm(const bigint &a, const bigint &b) {
return a / gcd(a, b) * b;
}
void read(const string &s) {
sign = 1;
a.clear();
int pos = 0;
while (pos < (int) s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-')
sign = -sign;
++pos;
}
for (int i = s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++)
x = x * 10 + s[j] - '0';
a.push_back(x);
}
trim();
}
friend istream& operator>>(istream &stream, bigint &v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream& operator<<(ostream &stream, const bigint &v) {
if (v.sign == -1)
stream << '-';
stream << (v.a.empty() ? 0 : v.a.back());
for (int i = (int) v.a.size() - 2; i >= 0; --i)
stream << setw(base_digits) << setfill('0') << v.a[i];
return stream;
}
static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
vector<long long> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < (int) p.size(); i++)
p[i] = p[i - 1] * 10;
vector<int> res;
long long cur = 0;
int cur_digits = 0;
for (int i = 0; i < (int) a.size(); i++) {
cur += a[i] * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int) cur);
while (!res.empty() && !res.back())
res.pop_back();
return res;
}
typedef vector<long long> vll;
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++)
a2[i] += a1[i];
for (int i = 0; i < k; i++)
b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int) a1b1.size(); i++)
r[i] -= a1b1[i];
for (int i = 0; i < (int) a2b2.size(); i++)
r[i] -= a2b2[i];
for (int i = 0; i < (int) r.size(); i++)
res[i + k] += r[i];
for (int i = 0; i < (int) a1b1.size(); i++)
res[i] += a1b1[i];
for (int i = 0; i < (int) a2b2.size(); i++)
res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint &v) const {
vector<int> a6 = convert_base(this->a, base_digits, 6);
vector<int> b6 = convert_base(v.a, base_digits, 6);
vll a(a6.begin(), a6.end());
vll b(b6.begin(), b6.end());
while (a.size() < b.size())
a.push_back(0);
while (b.size() < a.size())
b.push_back(0);
while (a.size() & (a.size() - 1))
a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < (int) c.size(); i++) {
long long cur = c[i] + carry;
res.a.push_back((int) (cur % 1000000));
carry = (int) (cur / 1000000);
}
res.a = convert_base(res.a, 6, base_digits);
res.trim();
return res;
}
};
void slove() {
bigint n; cin >> n;
cout << n*(n-1)/2 << '\n';
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
int t; cin >> t; while(t--)
slove();
return 0;
} J.最大值
思路：
没学过KMP，所以就暴力做咯
代码：
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define rep(i, a, n) for (int i=(a); i<(n); i++)
#define per(i, a, n) for (int i=(a); i>(n); i--)
typedef long long ll;
const int maxn = 2e5+5;
const int mod = 1e9+7;
using namespace std;
void slove() {
string s;
cin >> s;
vector<int> v;
ll l = s.size(), ans=0;
rep(i, 1, l) {
if (s[i] == s[0]) v.push_back(i);
}
for(auto c:v) {
ll cnt=1;
while(s[c+cnt] == s[cnt]) cnt++;
ans = max(cnt, ans);
}
cout << ans << '\n';
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
int t; cin >> t; while(t--)
slove();
return 0;
} 
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