A题
题意：
输入r, c，代表r*c的矩阵，接下来一行，是r个数，代表每一行里最大的数；接下来一行，是c个数，代表每一列中的最大数。求所给数据是否冲突。
思路：判断r个数中最大数maxr和c个数中最大数maxc是否相等，相等即能构造矩阵，否则不能。原因是maxr和maxc都是矩阵最大值，只有两数相等才可能构成矩阵。
代码：
1 #include<bits/stdc++.h> 2 3 #define vint vector<int> 4 #define vstr vector<string> 5 #define vll vector<long long> 6 #define ll long long 7 #define ull unsigned long long 8 #define pf printf 9 #define sf scanf 10 #define sfd(n) scanf("%d", &n) 11 #define sflf(n) scanf("%lf", &n) 12 #define sfll(n) scanf("%lld", &n) 13 #define pfd(n) printf("%d", n) 14 #define pflf(n) printf("%lf", n) 15 #define pfll(n) printf("%lld", n) 16 #define pft printf("\t") 17 #define pfn printf("\n") 18 #define pfk printf(" ") 19 #define PI 3.1415926 20 #define MAX 100000 21 22 using namespace std; 23 24 int n, m; 25 int a[105], b[105]; 26 27 int main() { 28 cin >> n >> m; 29 int maxr = -1, maxc = -1; 30 for (int i = 1; i <= n; i++) { 31 cin >> a[i]; 32 maxr = max(maxr, a[i]); 33 } 34 for (int i = 1; i <= m; i++) { 35 cin >> b[i]; 36 maxc = max(maxc, b[i]); 37 } 38 if( maxr==maxc ) { 39 cout << "possible" << endl; 40 }else { 41 cout << "impossible" << endl; 42 } 43 44 45 return 0; 46 }
F题
题意：
给定n， 且n = m^2 - k^2 ，求m，k。
思路：
设m = k+x；
则m^2 - k^2 = 2*x*k+x^2
即n = 2*x*k+x^2 , x、k是正整数
所以当x是奇数时，n为奇数；当x是偶数时，n是4的倍数；
所以
- n是奇数：x可以取1，即n = 2*k+1, m = k+1;
- n是4的倍数：x取2，即n = 4*k+4, m = k+2;
- n%2==0&&n%4!=0: 无法满足式子；
代码：
1 #include<bits/stdc++.h> 2 3 #define vint vector<int> 4 #define vstr vector<string> 5 #define vll vector<long long> 6 #define ll long long 7 #define ull unsigned long long 8 #define pf printf 9 #define sf scanf 10 #define sfd(n) scanf("%d", &n) 11 #define sflf(n) scanf("%lf", &n) 12 #define sfll(n) scanf("%lld", &n) 13 #define pfd(n) printf("%d", n) 14 #define pflf(n) printf("%lf", n) 15 #define pfll(n) printf("%lld", n) 16 #define pft printf("\t") 17 #define pfn printf("\n") 18 #define pfk printf(" ") 19 #define PI 3.1415926 20 #define MAX 100000 21 22 using namespace std; 23 24 int main() { 25 ll n, m, k; 26 bool f = true; 27 cin >> n; 28 if( n%2!=0 ) { 29 k = (n-1)/2; 30 m = k+1; 31 }else if( n%4==0 ){ 32 k = (n-4)/4; 33 m = k+2; 34 }else { 35 f = false; 36 } 37 if( f ) { 38 cout << m << " " << k; 39 }else { 40 pf("impossible"); 41 } 42 43 return 0; 44 }
I题
题意：
求公式最大值
思路：
sos保存前k项的平方和，sum保存后（n-k）项的和，
遍历k，求最大值
代码：
1 #include<bits/stdc++.h> 2 3 #define vint vector<int> 4 #define vstr vector<string> 5 #define vll vector<long long> 6 #define ll long long 7 #define ull unsigned long long 8 #define pf printf 9 #define sf scanf 10 #define sfd(n) scanf("%d", &n) 11 #define sfc(n) scanf("%c", &n) 12 #define sflf(n) scanf("%lf", &n) 13 #define sfll(n) scanf("%lld", &n) 14 #define pfd(n) printf("%d", n) 15 #define pfc(n) printf("%c", n) 16 #define pflf(n) printf("%lf", n) 17 #define pfll(n) printf("%lld", n) 18 #define pft printf("\t") 19 #define pfn printf("\n") 20 #define pfk printf(" ") 21 #define PI 3.1415926 22 #define MAX 100000 23 24 using namespace std; 25 26 int main() { 27 int n; 28 cin >> n; 29 int* num = new int[n]; 30 cin >> num[0]; 31 ll sos = num[0]*num[0], sum = 0, max = 0, ans; 32 for( int i=1; i<n; i++ ) { 33 cin >> num[i]; 34 sum += num[i]; 35 } 36 for( int k=1; k<n; k++ ) { 37 ans = sos*sum; 38 if( max<ans ) { 39 max = ans; 40 } 41 sos += num[k]*num[k]; 42 sum -= num[k]; 43 } 44 pfll(max); 45 46 return 0; 47 }
