PAT 1044 1044 Shopping in Mars （前缀和+二分）

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:

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

代码（前缀和、暴力）：

//前缀和的做法
#include 
#include 
using namespace std;

typedef pair PII;

const int N=100010;
int a[N],b[N];
int n,m;

int main(){
    vector v; 
    cin >> n >> m;
    
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        b[i]=b[i-1]+a[i];
    }
    
    int min=0x3f3f3f3f;
    for(int i=1;i<=n;i++){
        for(int j=0;j=m){
                if(b[i]-b[j]
#include 
using namespace std;

typedef pair PII;

const int N=100010;
int n,m;
int a[N];

int main(){
    vector v;
    int min=0x3f3f3f3f;
    cin >> n >> m;
    
    for(int i=1;i<=n;i++)    scanf("%d",&a[i]);
    
    for(int i=1;i<=n;i++){
        int sum=0;
        for(int j=i;j<=n;j++){
            sum+=a[j];
            if(sum>=m){
                if(sum

还是大佬的代码好使：

真的没有想到前缀和+二分这样的方法，忽略了一个乱序数组（每个元素都是正数）的前缀和是一个递增数组的特点！！！

#include 
#include 
using namespace std;

vector sum,res;
int n,m;

void func(int i,int &j,int &tempsum){
    int l=i,r=n;
    while(l>1;
        if(sum[mid]-sum[i-1]>=m) r=mid;
        else l=mid+1;
    }
    j=r;
    tempsum=sum[j]-sum[i-1];
}

int main(){
    cin >> n >> m;
    sum.resize(n+1);
    
    for(int i=1;i<=n;i++){
        scanf("%d",&sum[i]);
        sum[i]+=sum[i-1];
    }
    
    int minans=sum[n];//题目说明有足够的钱购买物品，所以赋值所有钻石之和肯定是大于等于账单价格
    for(int i=1;i<=n;i++){
        int j,tempsum;
        func(i,j,tempsum);
        if(tempsum>minans)  continue;
        if(tempsum>=m){
            if(tempsum

好好学习，天天向上！

我要考研！