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 F - String Partition

John was absurdly busy for preparing a programming contest recently. He wanted to create a ridiculously easy problem for the contest. His problem was not only easy, but also boring: Given a list of non-negative integers, what is the sum of them?

However, he made a very typical mistake when he wrote a program to generate the input data for his problem. He forgot to print out spaces to separate the list of integers. John quickly realized his mistake after looking at the generated input file because each line is simply a string of digits instead of a list of integers.

He then got a better idea to make his problem a little more interesting: There are many ways to split a string of digits into a list of non-zero-leading (0 itself is allowed) 32-bit signed integers. What is the maximum sum of the resultant integers if the string is split appropriately?

Input
The input begins with an integer N (≤ 500) which indicates the number of test cases followed. Each of the following test cases consists of a string of at most 200 digits.

Output
For each input, print out required answer in a single line.

Sample input

Sample output
1234554321
543211239
0
2121212124
214748372
5555555666

---

動態規劃

#include <stdio.h>
#include <string.h>
const long long limit = 2147483647LL;
long long max(long long x, long long y) {
    return x > y ? x : y;
}
int main() {
    int t, i, j;
    char str[201];
    scanf("%d", &t);
    while(t--) {
        scanf("%s", str);
        int slen = strlen(str), i;
        long long DP[202] = {}, tmp = 0;
        for(i = 0; i < slen; i++) {
            if(str[i] == '0') {
                DP[i+1] = max(DP[i+1], DP[i]);
            } else {
                tmp = 0;
                for(j = i; j < slen; j++) {
                    tmp = tmp*10 + str[j]-'0';
                    if(tmp > limit)
                        break;
                    DP[j+1] = max(DP[j+1], DP[i]+tmp);
                }
            }
        }
        printf("%lld\n", DP[slen]);
    }
    return 0;
}