PAT甲级 1044

Shopping in Mars

Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M).Whenmakingthepayment,thechaincanbecutatanypositionforonlyonceandsomeofthediamondsaretakenoffthechainonebyone.Onceadiamondisoffthechain,itcannotbetakenback.Forexample,ifwehaveachainof8diamondswithvaluesM3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:

1. Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).
2. Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
3. Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).
   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.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=105), the total number of diamonds on the chain, and M (<=108), the amount that the customer has to pay. Then the next line contains N positive numbers D1 … DN (Di<=103 for all i=1, …, N) which are the values of the diamonds. All the numbers in a line are separated by a space.

Output Specification:

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.

Sample Input 1:
16 15
3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13
Sample Output 1:
1-5
4-6
7-8
11-11
Sample Input 2:
5 13
2 4 5 7 9
Sample Output 2:
2-4
4-5

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1. 考点
   
   1. 二分法
2. 题解
   
   1. 用一个数组记录前面的和（因为是连续的，所以可以通过两个下标相减得到和）
   2. 遍历一个下标，另一个下标用二分法去找即可

#include<iostream>#include<stdio.h>#include<limits.h>#include<vector>using namespace std;const int maxn=100000+10;int sum[maxn];vector<pair<int,int> >vp;void binary(int low,int high,int M,int &total,int &idx){    int mid,start=low;    while(low<=high){        mid=(low+high)/2;        if(sum[mid]-sum[start]>=M){            high=mid-1;        }else{            low=mid+1;        }    }    idx=low;    total=sum[idx]-sum[start];}int main(){    freopen("./in.h","r",stdin);    int N,M,tmp;    scanf("%d %d",&N,&M);    for(int i=1;i<=N;++i){        scanf("%d",&tmp);        sum[i]=sum[i-1]+tmp;    }    int minNum=INT_MAX,idx;    for(int i=1;i<=N;++i){        binary(i-1,N,M,tmp,idx);        if(tmp<M)continue;        if(minNum>=tmp){            if(minNum>tmp) vp.clear();            minNum=tmp;            vp.push_back(make_pair(i,idx));        }    }    for(auto it=vp.begin();it!=vp.end();++it){        printf("%d-%d\n",it->first,it->second);    }    return 0;}