題單傳送門
成都創(chuàng)新互聯(lián)秉承實現(xiàn)全網價值營銷的理念,以專業(yè)定制企業(yè)官網,成都做網站、網站制作、成都外貿網站建設,小程序開發(fā),網頁設計制作,手機網站開發(fā),全網整合營銷推廣幫助傳統(tǒng)企業(yè)實現(xiàn)“互聯(lián)網+”轉型升級專業(yè)定制企業(yè)官網,公司注重人才、技術和管理,匯聚了一批優(yōu)秀的互聯(lián)網技術人才,對客戶都以感恩的心態(tài)奉獻自己的專業(yè)和所長。九進制轉十進制#include#include
#include#includeusing namespace std;
int main(){int a = 2022, cnt = 1, ans = 0;
while(a){ans += a % 10 * cnt;
cnt *= 9;
a /= 10;
}
cout<< ans<< endl;
return 0;
}
順子日期#include#include
#include#includeusing namespace std;
int main(){cout<< 14<< endl;
return 0;
}
刷題統(tǒng)計#include#include
#include#includeusing namespace std;
int main(){long long a, b, n;
cin >>a >>b >>n;
long long week = 5 * a + 2 * b;
long long ans = n / week * 7;
n %= week;
for(int i = 1; i<= 7; ++i){if(n<= 0){cout<< ans<< endl;
return 0;
}
switch(i){case 1:
case 2:
case 3:
case 4:
case 5:
n -= a;
break;
case 6:
case 7:
n -= b;
}
ans++;
}
cout<< ans<< endl;
return 0;
}
修剪灌木#include#include
#include#include#includeusing namespace std;
int main(){int n;
cin >>n;
vectorv;
for(int i = 1; i<= n / 2; ++i){int t = 2 * max(n - i, i);
cout<< t<< endl;
v.push_back(t);
}
if(n % 2 == 1) cout<< n - 1<< endl;
for(int i = v.size() - 1; i >= 0; --i){cout<< v[i]<< endl;
}
return 0;
}
X進制減法#include#include
#include#includeusing namespace std;
typedef long long LL;
const LL N = 1e5 + 10, MOD = 1000000007;
LL a[N], b[N];
int main(){int n, ma, mb;
cin >>n >>ma;
for(int i = ma - 1; i >= 0; --i) cin >>a[i];
cin >>mb;
for(int i = mb - 1; i >= 0; --i) cin >>b[i];
LL ans = 0;
for(int i = max(ma, mb) - 1; i >= 0; --i)
ans = (ans * max({1LL * 2 , a[i] + 1 , b[i] + 1}) + a[i] - b[i]) % MOD;
cout<< ans<< endl;
return 0;
}
統(tǒng)計子矩陣#include#include
#include#includeusing namespace std;
typedef long long LL;
const int N = 510;
LL s[N][N];
int main() {LL n, m, k;
cin >>n >>m >>k;
for (int i = 1; i<= n; ++i)
for (int j = 1; j<= m; ++j) {cin >>s[i][j];
s[i][j] += s[i - 1][j];
}
LL ans = 0;
for (int i = 1; i<= n; ++i)
for (int j = i; j<= n; ++j)
for (int l = 1, r = 1, sum = 0; r<= m; ++r) {sum += s[j][r] - s[i - 1][r];
while (sum >k) {sum -= s[j][l] - s[i - 1][l];
l++;
}
ans += r - l + 1;
}
cout<< ans<< endl;
return 0;
}
積木畫#include#include
#include#includeusing namespace std;
typedef long long LL;
const int N = 1e7 + 10, mod = 1000000007;
LL f[N][3];
int main() {int n;
cin >>n;
f[0][0] = 1;
for (int i = 1; i<= n; i++) {f[i][0] = (f[i - 1][0] + f[i - 1][1] + f[i - 1][2]) % mod;
if (i >= 2) { f[i][0] = (f[i][0] + f[i - 2][0]) % mod;
f[i][1] = (f[i - 2][0] + f[i - 1][2]) % mod;
f[i][2] = (f[i - 2][0] + f[i - 1][1]) % mod;
}
}
cout<< f[n][0] % mod<< endl;
return 0;
}
李白打酒加強版#include#include
#include#includeusing namespace std;
const int N = 200, mod = 1000000007;
int f[N][N][N];
void solve() {int n, m;
cin >>n >>m;
memset(f, 0, sizeof f);
f[0][0][2] = 1;
for (int i = 1; i< n + m; ++i)
for (int j = 0; j< m; ++j)
for (int k = 0; k<= m; ++k) { if (!(k & 1)) f[i][j][k] = (f[i][j][k] + f[i - 1][j][k >>1]) % mod;
if (j >0) f[i][j][k] = (f[i][j][k] + f[i - 1][j - 1][k + 1]) % mod;
}
cout<< f[n + m - 1][m - 1][1]<< endl;
}
int main() {int T;
cin >>T;
while (T--) {solve();
}
return 0;
}
砍竹子#include#include
#include#include#include#include#includeusing namespace std;
typedef pairPII;
void solve() {long long n, x;
cin >>n;
priority_queue, less>q;
for(int i = 1; i<= n; ++i){cin >>x;
if(x != 1) q.push({x, i});
}
long long ans = 0;
while(!q.empty() && q.top().first != 1){auto t = q.top();
q.pop();
long long cnt = 1, num = sqrt(t.first / 2 + 1);
if(num != 1) q.push({num, t.second});
while(!q.empty() && q.top().first == t.first && q.top().second == t.second - cnt){ cnt++;
auto temp = q.top();
q.pop();
if(num != 1) q.push({num, temp.second});
}
ans++;
}
cout<< ans<< endl;
}
int main() {solve();
return 0;
}
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本文名稱:2022年藍橋杯省賽C++B組題解-創(chuàng)新互聯(lián)
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