题解

18 条题解

  • 4
    @ 2017-12-22 14:56:29

    《初等数论》第三版 83页定理4

    设a1,a2及c均为正整数,gcd(a1,a2) = 1; 当 c >a1a2 - a1 - a2不定方程必有非负解。

    证明原书很详细,略

    定理5
    设a1,a2及c均为正整数,gcd(a1,a2) = 1; 当 c > a1a2不定方程必有正解。

  • 4
    @ 2017-11-12 17:51:43

    woc,考试时打了八十多行!!!
    #include<iostream>
    using namespace std;
    int main()
    {
    long long a,b,c;
    cin>>a>>b;
    c=a*b-a-b;
    cout<<c;
    return 0;
    }

  • 3
    @ 2018-01-29 21:55:12

    Accepted

    状态 耗时 内存占用

    #1 Accepted 2ms 256.0 KiB
    #2 Accepted 2ms 256.0 KiB
    #3 Accepted 2ms 256.0 KiB
    #4 Accepted 2ms 256.0 KiB
    #5 Accepted 2ms 256.0 KiB
    #6 Accepted 2ms 256.0 KiB
    #7 Accepted 2ms 256.0 KiB
    #8 Accepted 2ms 256.0 KiB
    #9 Accepted 2ms 256.0 KiB
    #10 Accepted 2ms 256.0 KiB
    #11 Accepted 2ms 256.0 KiB
    #12 Accepted 2ms 256.0 KiB
    #13 Accepted 2ms 256.0 KiB
    #14 Accepted 2ms 256.0 KiB
    #15 Accepted 1ms 256.0 KiB
    #16 Accepted 1ms 256.0 KiB
    #17 Accepted 2ms 256.0 KiB
    #18 Accepted 2ms 256.0 KiB
    #19 Accepted 2ms 256.0 KiB
    #20 Accepted 2ms 256.0 KiB
    水题!RP 2000+ 纪念!
    May The Father Of Understanding Guide U !

  • 2
    @ 2021-09-04 17:19:04

    十年OI一场空,不开long long见祖宗。

    #include <bits/stdc++.h>
    using namespace std;
    
    int main()
    {
        long long a,b;
        cin>>a>>b;
        cout<<a*b-a-b;
        return 0;
    }
    
  • 1
    @ 2024-08-27 17:31:29

    嗯,数论真不错
    嗯,这题真的水

    #include<iostream>
    using namespace std;
    int main(){
        long long a,b;
        cin >> a >> b;
        cout << a*b - a - b;
        return 0;
    }
    
  • 1
    @ 2018-03-23 01:40:46

    不懂数论,纯找规律

    #include <stdio.h>
    
    int main()
    {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%lld", (long long)a * b - a - b);
    }
    
    
  • 1
    @ 2018-02-06 10:26:36
    #include<stdio.h>
    int main(){
        unsigned long long int a, b;
        scanf("%lld%lld", &a, &b);
        printf("%lld", a*b-(a+b));
        return 0;
    }
    
  • 1
    @ 2017-12-07 20:45:50

    这题...呵呵

    #include<bits/stdc++.h>
    using namespace std;
    int main(){
        long long a,b;
        scanf("%lld%lld",&a,&b);
        printf("%lld",a*b-a-b);
        return 0;
    }
    
  • 1
    @ 2017-11-26 15:18:54

    公式
    #include <iostream>
    using namespace std;
    int main()
    {
    long long a,b;
    cin>>a>>b;
    cout<<a*b-a-b<<endl;
    return 0;
    }

  • 1
    @ 2017-11-14 19:01:34

    program lx;
    var a,b,c:qword;
    begin
    readln(a,b);
    c:=a*b-a-b;
    write(c);
    end.

  • 1
    @ 2017-11-12 19:17:45

    考试的时候写了半天暴力搜索
    var i,j:int64;
    begin
    readln(i,j);
    j:=(i-1)*(j-1)-1;
    writeln(j);
    end.

  • 0
    @ 2018-10-12 21:35:22

    #include<iostream>
    using namespace std;long long n,m;int main(){cin >> n >> m;cout << (long long)n*m-m-n << endl;return 0;}

  • 0
    @ 2017-11-13 17:33:28

    打表找的规律,,。,,。没用**任何**数学证明!!!
    详见我的blog: http://von-hyou.pub

    #include <cstdio>
    #include <cctype>
    #include <algorithm>
    
    typedef long long LL;
    
    inline LL getInt() {
      LL sum(0); char _c;
      while (!isdigit(_c = getchar()));
      do sum = sum * 10 + _c - '0';
      while (isdigit(_c = getchar()));
      return sum;
    }
    
    int main() {
      LL n = getInt(), m = getInt();
      if (n > m) std::swap(n, m);
      if (n & 1 && m & 1)
        printf("%lld\n", ((m >> 1) * (n >> 1) << 2) - 1);
      else {
        LL tmp = n & 1 ? n : m;
        n = n & 1 ? m : n, m = tmp;
        printf("%lld\n", (((n - 2) << 1) + 2) * (m >> 1) - 1);
      } return 0;
    }
    
  • 0
    @ 2017-11-13 10:34:36
    #include <cmath>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <iomanip>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <deque>
    #include <set>
    #include <limits>
    #include <string>
    #include <sstream>
    using namespace std;
    
    const int oo_min=0xcfcfcfcf,oo_max=0x3f3f3f3f;
    
    long long a,b;
    
    int main()
    {
        while (~scanf("%lld%lld",&a,&b))
            printf("%lld\n",a*b-a-b);
    }
    
    • @ 2017-11-13 10:35:10

      看出來的30s打完

  • 0
    @ 2017-11-12 17:10:44

    考试的时候 我居然用公式去验证我的暴力 还忘了longlong
    #include <cstdio>
    #include <iostream>
    using namespace std;
    long long int a,b;
    int main()
    {
    cin>>a>>b;
    cout<<a*b-a-b<<endl;
    return 0;
    }

  • 0
    @ 2017-11-11 19:04:52
    #include <iostream>
    using namespace std;
    
    long long a,b;
    int main()
    {
        cin >>a>>b;
        cout <<a*b-a-b<<endl;
    }
    
  • 0
    @ 2017-11-11 18:53:20

    #include <iostream>
    #include <cstdlib>
    #include <cstdio>

    using namespace std;

    long long a, b;
    long long ret;
    int main() {

    cin >> a >> b;

    ret = a* b - (a + b);
    if (ret < 0) {
    ret = 0;
    }
    cout << ret;
    system("pause");
    return 0;
    }

  • -1
    @ 2017-11-12 17:03:58

    傻逼数学结论题,不想说啥了
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <cmath>
    #include <cstdlib>
    using namespace std;
    long long a,b;
    long long ans;
    long long read()
    {
    char c;int ch=0,f=1;
    while(c<'0'||c>'9')
    {
    if(c=='-')
    f=-1;
    c=getchar();
    }
    while(c>='0'&&c<='9')
    {
    ch=ch*10+c-'0';
    c=getchar();
    }
    return ch*f;
    }
    int main()
    {
    a=read();b=read();
    ans=a*b-a-b;
    printf("%lld\n",ans);
    return 0;
    }

  • 1

信息

ID
2028
难度
2
分类
(无)
标签
递交数
965
已通过
433
通过率
45%
被复制
12
上传者