“强智杯"2020年湖南省大学生计算机程序设计竞赛
oj: 链接
A.2020
oj: 链接
题意
Bobo有一个只包含数字 0，1， 2 的，长度为 n 的字符串 s1 ... sn。他想选出最多的互不重叠的连续子串，这些子串都是2020。求最多可以选出的子串数量。
形式化的，他想求出最大的 k ，是的存在 k 个下标 i1，...，ik 满足 
题解
由于数据过小，直接暴力即可
代码
#include <bits/stdc++.h>
using namespace std;
#define _for(i, a) for(int i = 0, len = a; i < len; ++i)
#define _rep(i, a, b) for(int i = a, len = b; i <= len; ++i)
#define outval(a) cout << "debuging|" << #a << ":" << a << "\n";
#define dg if(debug)
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 100005;
int debug = 0;
inline LL read() {
LL x(0), f(1); char ch(getchar());
while(ch < '0' || ch > '9') {
f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
struct poi {
};
char s[maxn];
int n;
void init() {
}
int sol() {
init();
scanf("%s", s);
int ans = 0;
int n = strlen(s);
for(int i = 0; i + 3 < n;) {
if(s[i] == '2' && s[i + 1] == '0' && s[i + 2] == '2' && s[i + 3] == '0') {
++ans;
i += 4;
}
else ++i;
}
printf("%d\n", ans);
return 0;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
debug = 1;
#endif
while(cin>>n) {
sol();
}
return 0;
}
B.2020 vs 2018
oj: 链接
题意
Bobo有一副 n 行 m 列，只包含 '.' 和 'o' 两种字符的字符画，其中 '.' 是背景，'o' 画出了2018或者2020。请识别给出的字符画是2018还是2020的。
题解
只需要判断是否存在由'0'组成的1即可
代码
#include <bits/stdc++.h>
using namespace std;
#define _for(i, a) for(int i = 0, len = a; i < len; ++i)
#define _rep(i, a, b) for(int i = a, len = b; i <= len; ++i)
#define outval(a) cout << "debuging|" << #a << ":" << a << "\n";
#define dg if(debug)
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int debug = 0;
inline LL read() {
LL x(0), f(1); char ch(getchar());
while(ch < '0' || ch > '9') {
f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
struct poi {
};
int n, m;
char s[57][57];
void init() {
_rep(i,1,n) scanf("%s",s[i]+1);
_rep(i,1,n){
dg outval(i);
_rep(j,i,n){
dg outval(j);
_rep(k,1,m){
int f=1;
if(i>1&&s[i-1][k]=='o') continue;
if(j<n&&s[j+1][k]=='o') continue;
_rep(l,i,j){
if(s[l][k]!='o'){
f=0;
break;
}
if(k>1){
if(s[l][k-1]=='o'){
f=0;
break;
}
}
if(k<m){
if(s[l][k+1]=='o'){
f=0;
break;
}
}
}
if(f){
puts("2018");
return ;
}
}
}
}
_rep(i,1,m){
_rep(j,i,m){
_rep(k,1,n){
int f=1;
if(i>1&&s[k][i-1]=='o') continue;
if(j<m&&s[k][j+1]=='o') continue;
_rep(l,i,j){
if(s[k][l]!='o'){
f=0;
break;
}
if(k>1){
if(s[k-1][l]=='o'){
f=0;
break;
}
}
if(k<n){
if(s[k+1][l]=='o'){
f=0;
break;
}
}
}
if(f){
puts("2018");
return ;
}
}
}
}
puts("2020");
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
// debug = 1;
#endif
while(~scanf("%d%d",&n,&m)) init();
return 0;
}
D.String Commutativity
oj: 链接
题意
有n个字符串s1，...，sn，他想找到对数i <j，其中si + sj = sj + si。注意，a + b表示字符串a和b的串联，即，首先写字符串a，然后写字符串b。例abcd+efg=abcdefg。
题解
如果两个串有相同的最短循环节，则这两个串符合题目要求。用kmp求最短循环节，但kmp求出的最短循环节不一定符合题意，例如aba的最短循环节为ab，而kmp求出的为aba，所以应排除这种情况。即用m%x!=0判断是否最后一个循环节是否完全循环
代码
#include <cstring>
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <map>
using namespace std;
const int maxn=1e6+7;
const int INF=0x3f3f3f3f;
#define M 50015
int next1[maxn];
char str[maxn],mo[maxn];
int ans;
map<string,int> mp;
string tran(int x){
string a;
for(int i=0;i<x;i++){
a.push_back(mo[i]);
}
return a;
}
void getnext()
{
int i=0,j=-1,m=strlen(mo);
while(i<m){
if(j==-1||mo[i]==mo[j])
next1[++i]=++j;
else
j=next1[j];
}
}
int main()
{
int t;
while(~scanf("%d",&t)){
mp.clear();
long long sum=0;
while(t--){
ans=0;
scanf("%s",mo);
next1[0]=-1;
getnext();
int m=strlen(mo);
int x=m-next1[m];
if(x<m&&m%x!=0){
x=m;
}
//printf("x:%d\n", x);
string y=tran(x);
mp[y]++;
}
map<string,int>::iterator it;
it=mp.begin();
while(it!=mp.end()){
int n=it->second;
sum+=1LL*n*(n-1)/2;
it++;
}
printf("%lld\n",sum);
}
return 0;
}
G.奇矩阵
oj: 链接
题意
题解
队友写的签到题
代码
#include <bits/stdc++.h>
using namespace std;
#define _for(i, a) for(int i = 0, len = a; i < len; ++i)
#define _rep(i, a, b) for(int i = a, len = b; i <= len; ++i)
#define outval(a) cout << "debuging|" << #a << ":" << a << "\n";
#define dg if(debug)
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int debug = 0;
inline LL read() {
LL x(0), f(1); char ch(getchar());
while(ch < '0' || ch > '9') {
f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
struct poi {
};
int n, m;
int num[2];
void init() {
num[0] = num[1] = 0;
}
int sol() {
init();
_for(i, n) {
LL sum = 0;
_for(j, m) sum += read();
++num[sum & 1];
}
if(num[0] > 1 || num[1] > 1) return 0;
return 1;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
debug = 1;
#endif
while(cin>>n>>m) {
printf("%s\n", sol() ? "Yes":"No");
}
return 0;
}
H.矩阵并
oj: 链接
题意
题解
涛神过的题，完全看不懂
代码
#include <bits/stdc++.h>
using namespace std;
#define _for(i, a) for(int i = 0, len = a; i < len; ++i)
#define _rep(i, a, b) for(int i = a, len = b; i <= len; ++i)
#define outval(a) cout << "debuging|" << #a << ":" << a << "\n";
#define dg if(debug)
typedef long long LL;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1005;
const LL mod = 1000000007;
int debug = 0;
inline LL read() {
LL x(0), f(1); char ch(getchar());
while(ch < '0' || ch > '9') {
if(ch == '-') f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
struct poi {
};
LL a, b;
void exgcd(LL a, LL b, LL &g, LL &x, LL &y) {
if (!b) {
g = a;
x = 1;
y = 0;
} else {
exgcd(b, a % b, g, y, x);
y -= x * (a / b);
}
}
LL inv(LL x, LL mod) {
LL a, k, g;
exgcd(x, mod, g, a, k);
if(a < 0) a += mod;
return a;
}
LL getval(LL a, LL b) {
LL ans = 0;
ans += a * a % mod * b % mod * b;
ans += a * b % mod * b % mod;
ans += a * a % mod * b % mod;
ans += a * b % mod;
ans %= mod;
ans *= inv(4, mod);
return ans % mod;
}
int sol() {
LL x = read(), xx = read(), y = read(), yy = read();
LL ans = 0;
ans += getval(a, b);
dg printf("累积：%lld\n", getval(a, b));
LL w1 = max(0LL, min(b, yy) - y);
LL box1 = w1 * (w1 + 1) % mod * inv(2, mod) % mod * max(0LL, a - xx) % mod * (xx - x) % mod;
LL w2 = max(0LL, min(a, xx) - x);
if(a < x) w2 = 0;
dg printf("w1:%lld\tw2:%lld\n", w1, w2);
LL box2 = w2 * (w2 + 1) % mod * inv(2, mod) % mod * max(0LL, b - yy) % mod * (yy - y) % mod;
LL box3 = max(0LL, a - xx) * max(0LL, b - yy) % mod * (xx - x) % mod * (yy - y) % mod;
LL ans2 = box1 + box2 + box3 + getval(max(0LL, min(a, xx) - x), max(0LL, min(b, yy) - y)); ans2 %= mod;
dg printf("box1:%lld\tbox2:%lld\tbox3:%lld\n", box1, box2, box3);
dg printf("减去：%lld\n", ans2);
ans += mod - ans2;
LL ans3 = (xx - x) * (yy - y) % mod * a % mod * b % mod;
ans += ans3;
dg printf("额外：%lld\n", ans3);
ans %= mod;
printf("%lld\n", ans);
return 0;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
// debug = 1;
#endif
while(~scanf("%lld%lld", &a, &b)) {
sol();
}
return 0;
}
I.共线点
oj: 链接
题意
题解
模板题
代码
#include <bits/stdc++.h>
using namespace std;
#define _for(i, a) for(int i = 0, len = a; i < len; ++i)
#define _rep(i, a, b) for(int i = a, len = b; i <= len; ++i)
#define outval(a) cout << "debuging|" << #a << ":" << a << "\n";
#define dg if(debug)
typedef long long LL;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1005;
int debug = 0;
inline LL read() {
LL x(0), f(1); char ch(getchar());
while(ch < '0' || ch > '9') {
f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
struct poi {
int a, b, y;
bool operator<(const poi &b) const {
if(y != b.y) return y < b.y;
else if(a != b.a) return a < b.a;
else return this->b < b.b;
}
};
poi a[3];
void init() {
}
int sol() {
_rep(i, 1, 2) {
int aa, bb, yy;
scanf("%d%d%d", &aa, &bb, &yy);
a[i] = {aa, bb, yy};
}
sort(a, a + 3);
double x1 = 1.0 * (a[2].a - a[0].a) / (a[2].y - a[0].y) * (a[1].y - a[0].y) + a[0].a;
double x2 = 1.0 * (a[2].b - a[0].b) / (a[2].y - a[0].y) * (a[1].y - a[0].y) + a[0].b;
dg printf("xx1:%.2f\txx2:%.2f\n", x1, x2);
if(x1 - eps <= a[1].b && x2 + eps >= a[1].a) return 1;
return 0;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
debug = 1;
#endif
int aa, bb, yy;
while(~scanf("%d%d%d", &aa, &bb, &yy)) {
a[0] = {aa, bb, yy};
printf("%s\n", sol () ?"Yes":"No");
}
return 0;
}
