Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B

Pay the Price

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

 

In ancient days there was a country whose people had very interesting habits. Some of them were lazy, some were very rich, some were very poor and some were miser. Obviously, some of the rich were miser (A poor was never miser as he had little to spend) and lazy but the poor were lazy as well (As the poor were lazy they remained poor forever). The following things were true for that country

 

a)      As the rich were miser, no things price was more than 300 dollars (Yes! their currency was dollar).

b)      As all people were lazy, the price of everything was integer (There were no cents and so beggars always earned at least one dollar)

c)      The values of the coins were from 1 to 300 dollars, so that the rich (who were idle) could pay any price with a single coin.

 

Your job is to find out in how many ways one could pay a certain price using a limited number of coins (Note that the number of coins paid is limited but not the value or source. I mean there was infinite number of coins of all values). For example, by using three coins one can pay six dollars in 3 ways, 1+1+4, 1+2+3, and 2+2+2. Similarly, one can pay 6 dollars using 6 coins or less in 11 ways.

 

Input

The input file contains several lines of input. Each line of input may contain 1, 2 or 3 integers. The first integer is always N (0<=N<=300), the dollar amount to be paid. All other integers are less than 1001 and non-negative.

 

Output

For each line of input you should output a single integer.

 

When there is only one integer N as input, you should output in how many ways N dollars can be paid.

 

When there are two integers N and L1 as input, then you should output in how many ways N dollars can be paid using L1 or less coins.

 

When there are three integers N, L1 and L2 as input, then you should output in how many ways N dollars can be paid using L1, L1+1 …, L2 coins (summing all together). Remember that L1 is not greater than L2.

 

Sample Input

6

6 3

6 2 5

6 1 6

 

Sample Output

11

7

9

11

---

時間都耗在輸入 ...

#include <stdio.h>
#include <iostream>
#include <sstream>
using namespace std;

int main() {
    long long DP[301][301] = {};
    int i, j, k;
    char str[1000];
    DP[0][0] = 1;
    for(i = 1; i <= 300; i++) {
        for(j = i; j <= 300; j++) {
            for(k = 1; k <= j; k++)
                DP[j][k] += DP[j-i][k-1];
        }
    }
    while(gets(str)) {
        stringstream sin(str);
        int num[3], n = 0;
        while(sin >> num[n])
            n++;
        int p, q;
        if(n == 1)   
            p = 0, q = num[0];
        else if(n == 2)
            p = 0, q = num[1];
        else
            p = num[1], q = num[2];
        n = num[0];
        if(q > n)    q = n;
        long long ans = 0;
        for(i = p; i <= q; i++)
            ans += DP[n][i];
        printf("%lld\n", ans);
    }
    return 0;
}