PAT-A-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$). 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:

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).
Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).
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 (≤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.

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
代码如下：

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=100010;
int sum[maxn];
//查找第一个大于x的位置
int upper_bound(int left,int right,int k){
    int mid;
    while(left<right){
        mid=(left+right)/2;
        if(sum[mid]>k){
            right=mid;
        }else{
            left=mid+1;
        }
    }
    return left;
}
int main(){
    int n,m,a[maxn],nearM=100000010;
    sum[0]=0;
    scanf("%d%d",&n,&m);//元素个数 和值
    for(int i = 1;i <= n;i++){
        scanf("%d",&a[i]);
        sum[i]=sum[i-1]+a[i];
    }
    // for(int i = 1;i <=n ;i++){
    //     cout<
    // }
    //第一次遍历求出大于等于m的最近接m的值nearM
    for(int i = 1;i <=n ;i++){//枚举左端点
        int j=upper_bound(i,n+1,sum[i-1]+m);//j是大于查找值的第一个数的下标
        // cout<
        if(sum[j-1]-sum[i-1]==m){ //查找成功
            nearM=m;//最接近和值的就是m 
            break;
        }else if(j<=n&&sum[j]-sum[i-1]<nearM){//往后多加一位
            //存在大于m的解并小于nearM
            nearM=sum[j]-sum[i-1];//更新当前nearM
        }
    }
    //第二次遍历找那些和值恰好为nearM的方案输出
    for(int i = 1;i <=n ;i++){
        int j=upper_bound(i,n+1,sum[i-1]+nearM);
        // cout<
        if(sum[j-1]-sum[i-1]==nearM){//查找成功
            printf("%d-%d\n",i,j-1);
        }
    }

    return 0;
}