PAT 1033Shopping in Mars (25)（子序列）

题目

1044. Shopping in Mars (25)时间限制100 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueShopping 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 1:16 153 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13Sample Output 1:1-54-67-811-11Sample Input 2:5 132 4 5 7 9Sample Output 2:2-44-5

解题思路

• 1.如果sum[i](只从开始到第i个累加和)<m,那么它的字数列也会小于，则不用考虑。
• 2.如果sum[i](只从开始到第i个累加和)>m,那么我们循环减掉前面的，直到它等于m，你就找到答案了。
• 3.如果找不到，可在上述第二步中记录一个最小值。

代码

#include<iostream>#include<cstdio>using namespace std;int main(int argc, char *argv[]){    int n,m,sum[100000+5]={0},tem,first=0,mindiff = 100000000+5;    bool isFind = false;    scanf("%d %d",&n,&m);    for (int i = 1; i <= n; ++i) {        scanf("%d",&tem);        sum[i]=sum[i-1]+tem;        //如果``sum[i](只从开始到第i个累加和)>m``,那么我们循环减掉前面的，直到它等于`m`，你就找到答案了。        while (sum[i] - sum[first]>m) {            mindiff = min(mindiff,sum[i]-sum[first]);            first++;        }        if (sum[i] - sum[first] == m) {            printf("%d-%d\n",first+1,i);            isFind = true;        }    }    //如果找不到，可在上述第二步中记录一个最小值。然后再类似第二部求解。    if (!isFind) {        int j=0;        for (int i = 1; i <= n; ++i) {            while (sum[i] - sum[j]>mindiff) {                j++;            }            if (sum[i] - sum[j] == mindiff) {                printf("%d-%d\n",j+1,i);            }        }    }    return 0;}