2019.3.14解题报告&补题报告

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：

#include<bits/stdc++.h>

#define vint vector<int>
#define vstr vector<string>
#define vll vector<long long>
#define ll long long
#define ull unsigned long long
#define pf printf
#define sf scanf
#define sfd(n) scanf("%d", &n)
#define sflf(n) scanf("%lf", &n)
#define sfll(n) scanf("%lld", &n)
#define pfd(n) printf("%d", n)
#define pflf(n) printf("%lf", n)
#define pfll(n) printf("%lld", n)
#define pft    printf("	")
#define pfn printf("
")
#define pfk printf(" ")
#define PI 3.1415926
#define MAX 100000

using namespace std;

int n, m;
int a[105], b[105];

int main() {
    cin >> n >> m;
    int maxr = -1, maxc = -1;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        maxr = max(maxr, a[i]);
    }
    for (int i = 1; i <= m; i++) {
        cin >> b[i];
        maxc = max(maxc, b[i]);
    }
    if( maxr==maxc ) {
        cout << "possible" << endl;
    }else {
        cout << "impossible" << endl;
    }
         

    return 0;
}

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：

#include<bits/stdc++.h>

#define vint vector<int>
#define vstr vector<string>
#define vll vector<long long>
#define ll long long
#define ull unsigned long long
#define pf printf
#define sf scanf
#define sfd(n) scanf("%d", &n)
#define sflf(n) scanf("%lf", &n)
#define sfll(n) scanf("%lld", &n)
#define pfd(n) printf("%d", n)
#define pflf(n) printf("%lf", n)
#define pfll(n) printf("%lld", n)
#define pft    printf("	")
#define pfn printf("
")
#define pfk printf(" ")
#define PI 3.1415926
#define MAX 100000

using namespace std;

int main() {
    ll n, m, k;
    bool f = true;
    cin >> n;
    if( n%2!=0 ) {
        k = (n-1)/2;
        m = k+1;
    }else if( n%4==0 ){
        k = (n-4)/4;
        m = k+2;
    }else {
        f = false;
    }
    if( f ) {
        cout << m << " " << k;
    }else {
        pf("impossible");
    }

    return 0;
}

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：

#include<bits/stdc++.h>

#define vint vector<int>
#define vstr vector<string>
#define vll vector<long long>
#define ll long long
#define ull unsigned long long
#define pf printf
#define sf scanf
#define sfd(n) scanf("%d", &n)
#define sfc(n) scanf("%c", &n)
#define sflf(n) scanf("%lf", &n)
#define sfll(n) scanf("%lld", &n)
#define pfd(n) printf("%d", n)
#define pfc(n) printf("%c", n)
#define pflf(n) printf("%lf", n)
#define pfll(n) printf("%lld", n)
#define pft    printf("	")
#define pfn printf("
")
#define pfk printf(" ")
#define PI 3.1415926
#define MAX 100000

using namespace std;

int main() {
    int n;
    cin >> n;
    int* num = new int[n];
    cin >> num[0];
    ll sos = num[0]*num[0], sum = 0, max = 0, ans;
    for( int i=1; i<n; i++ ) {
        cin >> num[i];
        sum += num[i];
    }
    for( int k=1; k<n; k++ ) {
        ans = sos*sum;
        if( max<ans ) {
            max = ans;
        }
        sos += num[k]*num[k];
        sum -= num[k];
    }
    pfll(max);

    return 0;
}