18 条题解
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4冲啊小笼包 LV 9 @ 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。 -
42017-11-12 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;
} -
32018-01-29 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 ! -
22021-09-04 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; }
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12024-08-27 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; }
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12018-03-23 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); }
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12018-02-06 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; }
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12017-12-07 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; }
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12017-11-26 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;
} -
12017-11-14 19:01:34@
program lx;
var a,b,c:qword;
begin
readln(a,b);
c:=a*b-a-b;
write(c);
end. -
12017-11-12 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. -
02018-10-12 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;} -
02017-11-13 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; }
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02017-11-13 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); }
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02017-11-12 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;
} -
02017-11-11 19:04:52@
#include <iostream> using namespace std; long long a,b; int main() { cin >>a>>b; cout <<a*b-a-b<<endl; }
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02017-11-11 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;
} -
-12017-11-12 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
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