标签：problem mount play 函数 test html Fix htm note
1044 Shopping in Mars（25 分）
Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:
Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.
If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.
Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤10?5??), the total number of diamonds on the chain, and M (≤10?8??), the amount that the customer has to pay. Then the next line contains N positive numbers D?1???D?N?? (D?i??≤10?3?? for all i=1,?,N) which are the values of the diamonds. All the numbers in a line are separated by a space.
For each test case, print i-j in a line for each pair of i ≤ j such that Di + ... + Dj = M. Note that if there are more than one solution, all the solutions must be printed in increasing order of i.
If there is no solution, output i-j for pairs of i ≤ j such that Di + ... + Dj >M with (Di + ... + Dj ?M) minimized. Again all the solutions must be printed in increasing order of i.
It is guaranteed that the total value of diamonds is sufficient to pay the given amount.
16 15
3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13
1-5
4-6
7-8
11-11
5 13
2 4 5 7 9
2-4
4-5题目大意：给出一串数表示钻石的价值，并且给出m，如果有一连串钻石价值正好等于m，那么就输出这一串；如果没有正好相等的和，那么就输出那些>m，并且和m差值最小的！
//第一次提交这样，就是通过两层循环来做，
#include <iostream> #include <vector> #include <cmath> using namespace std; int dio[100001]; int main() { int n,m; cin>>n>>m; for(int i=0;i<n;i++){ cin>>dio[i]; } bool flag=false; for(int i=0;i<n;i++){ int sum=0; for(int j=i;j<n;j++){ sum+=dio[j]; if(sum==m){ flag=true; cout<<i+1<<‘-‘<<j+1<<‘\n‘;break; } } } vector<int> vt; if(!flag){//如果没有解。 int mini=100001; for(int i=0;i<n;i++){ int sum=0; for(int j=i;j<n;j++){ sum+=dio[j]; if(sum>m&&sum-m<mini){ vt.clear(); vt.push_back(i+1); vt.push_back(j+1); //cout<<i+1<<"="<<j+1<<" "<<sum<<‘\n‘; mini=sum-m;break; }else if(sum>m&&sum-m==mini){ vt.push_back(i+1); vt.push_back(j+1); // cout<<i+1<<"+"<<j+1<<‘\n‘; break; } } } } for(int i=0;i<vt.size();i+=2){ cout<<vt[i]<<‘-‘<<vt[i+1]<<‘\n‘; } return 0; }
//但是提交到牛客网上，
运行超时:您的程序未能在规定时间内运行结束，请检查是否循环有错或算法复杂度过大。
case通过率为40.00%
//顿时感觉凉凉。提交到pat上，2，3，5测试点运行超时。
代码来自:https://www.liuchuo.net/archives/2939
#include <iostream> #include <vector> using namespace std; vector<int> sum, resultArr; int n, m; void Func(int i, int &j, int &tempsum) {//这个求的是从i到j的和。 //表面上是left，但是while里的if条件仍然只是>=m int left = i, right = n; while(left < right) { int mid = (left + right) / 2; if(sum[mid] - sum[i-1] >= m) right = mid;//必须是=mid，如果是mid-1，那很有可能和就不够了。 else left = mid + 1; } j = right; tempsum = sum[j] - sum[i-1]; } int main() { scanf("%d%d", &n, &m); sum.resize(n+1); for(int i = 1; i <= n; i++) { scanf("%d", &sum[i]); sum[i] += sum[i-1];//这是前缀和？ } int minans = sum[n]; for(int i = 1; i <= n; i++) { int j, tempsum; Func(i, j, tempsum);//直接传引用。 if(tempsum > minans) continue;//感觉这一句不用吧，肯定不会大于的。 //明白了，这个在i=1第一次循环的时候是没用，但是minans更新之后就有用了。 //判断>当前的都不选择。 if(tempsum >= m) { if(tempsum < minans) { resultArr.clear(); minans = tempsum; } resultArr.push_back(i); resultArr.push_back(j); } } for(int i = 0; i < resultArr.size(); i += 2) printf("%d-%d\n", resultArr[i], resultArr[i+1]); return 0; }
//学习了
1.当程序正常超时时，需要考虑二分法。
2.使用引用传值，减少函数调用时间。
3.这里的mid求的不是到left的和，而是到i的和！和往常的不一样。
//这个二分我感觉确实挺难的，相不到，
代码来自：https://www.cnblogs.com/chenxiwenruo/p/6677802.html
#include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <vector> #define INF 0x3f3f3f3f using namespace std; const int maxn=100000+5; int n,m; int diamond[maxn]; struct Ans{ int i,j; }; vector<Ans> ans; int main() { scanf("%d %d\n",&n,&m); for(int i=1;i<=n;i++) scanf("%d",&diamond[i]); int minlost=INF; Ans tmp; int sum=0; int start=1; for(int i=1;i<=n;i++){ sum+=diamond[i]; //printf("%d %d %d %d %d\n",start,i,sum,sum-m,minlost); if(sum<m) continue; if(sum-m<=minlost){ if(sum-m<minlost){ ans.clear(); minlost=sum-m; } tmp.i=start; tmp.j=i; ans.push_back(tmp); } if(sum>=m && start<i){//这里是在做什么？ sum-=diamond[i];//明白了，-钻石i的值，和钻石start的值， i--;//如果减去diamond[i]，并且减去start的值，那么总和肯定小于m，那么 sum-=diamond[start];//肯定要就着i往下加了。 start++;//目前的sum肯定达不到m，所以肯定需要往后加。 printf("%d\n",sum); } } for(int i=0;i<ans.size();i++){ printf("%d-%d\n",ans[i].i,ans[i].j); } return 0; }
//这个也写得很好，但是我有点不太理解。
// 但是接着说上述二分方法的话，也就是原来二分，求和之后二分的方法。
PAT 1044 Shopping in Mars[二分][难]
标签：problem mount play 函数 test html Fix htm note
原文地址：https://www.cnblogs.com/BlueBlueSea/p/9557412.html
