1 题目描述（题目链接）

2 题解

由题意得，na=m+c−bna=m+c-bna=m+c−b，则na∈[m+l−r,m+r−l]na\in [m+l-r,m+r-l]na∈[m+l−r,m+r−l]。对于a∈[l,r]a\in [l,r]a∈[l,r]，遍历求解n=⌊m+r−la⌋n=\lfloor \cfrac{m+r-l}{a} \rfloorn=⌊am+r−l⌋，并验证n∗an*an∗a是否在上述范围内，如果在，则b,cb,cb,c的值只需要满足c−b=na−mc-b = na-mc−b=na−m即可。

#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;int main(int argc, const char * argv[]) {// insert code here...int t;cin>>t;ll l[20], r[20], m[20];ll i, a, n;ll left, right;for (i = 0; i < t; i++){cin>>l[i]>>r[i]>>m[i];}for (i = 0; i < t; i++){left = l[i] - r[i] + m[i];right = r[i] - l[i] + m[i];for (a = l[i]; a <= r[i]; a++){n = floor(right/a);if (left <= a*n && a*n <= right){break;}}if (a*n - m[i] < 0)cout<<a<<" "<<r[i]<<" "<<r[i]+a*n-m[i]<<endl;elsecout<<a<<" "<<r[i]+m[i]-a*n<<" "<<r[i]<<endl;}return 0;
}

t = int(input())l, r, m = [], [], []for i in range(t):lrm = input().split()l.append(int(lrm[0]))r.append(int(lrm[1]))m.append(int(lrm[2]))for i in range(t):left = m[i] + l[i] - r[i]right = m[i] + r[i] - l[i]for a in range(l[i], r[i] + 1):n = right//aif left <= a*n <= right:breakif a*n < m[i]:print(a, r[i], r[i]+a*n-m[i], end=" ")else:print(a, r[i]+m[i]-a*n, r[i], end=" ")

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Codeforces 1379B.Dubious Cyrpto

1 题目描述（题目链接）

2 题解

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