ACM-ACMICPC (AC, dynamic planning/traversing all strings, three pieces of code)

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• Description:
• Input:
• Output:
• Sample Input:
• Sample Output:

ACMICPC
Time Limit: 1000 MS Memory Limit: 32768 K

Description:

The uppercase letter A-Z corresponds to an integer [-13, 12], so a string corresponds to an integer column. The sum of the integer columns corresponding to the string is called the characteristic value of the string. For example, if the integer column {-13,-11,-1} corresponding to the string ACM is set to (-13) + (-11) + (-1) =-25; the feature value of its sub-string AC is-24; the feature value of its sub-string M is-1. For a string, find the maximum feature value of all the substrings of the string.

Input:

The positive integer N (1 <= N <= 1000) in the first line indicates that there are N groups of test data, each of which is composed of a string. A string can only consist of uppercase letters and A-Z and cannot exceed 1000 characters in length.

Output:

There are N rows of output, and each row is the output result of a group of test data.

Sample Input:

2ACMANGRY

Sample Output:

-115

Http://acm.zjut.edu.cn/ShowProblem.aspx? ShowID = 1021

 

/* ------------------------------------------------------------------------------ Non-dynamic planning solution: Practice: traverse all strings to find the largest one. Time: 297MS memory: 196 K -------------------------------------------------------------------------------- */# include <stdio. h> char szStr [1004]; int a [1004]; int main (void) {int I = 0; int j = 0; int t = 0; int n = 0; int nMax =-1000; scanf ("% d", & t); getchar (); while (t --> 0) {gets (szStr); nMax =-1000; j = 1; I = 0; while (szStr [I]! = '\ 0') {a [j ++] = szStr [I]-65-13; I ++;} n = j-1; a [0] = 0; for (I = 1; I <= n; ++ I) {if (a [I]> = nMax) {nMax = a [I];} a [I] + = a [I-1]; if (a [I]> = nMax) {nMax = a [I];} for (j = 1; j <I; ++ j) {if (a [I]-a [j]> = nMax) {nMax = a [I]-a [j] ;}} printf ("% d \ n", nMax) ;}return 0 ;}

 

/* ------------------------------------- Dynamic planning solution: Time: 9 ms memory: 200 K ----------------------------------------- */# include <stdio. h ># define max (x, y) (x)> (y )? (X): (y) char szStr [1004]; int a [1004]; int B [1004]; int main (void) {int I = 0; int j = 0; int t = 0; int n = 0; int nMax =-1000; scanf ("% d", & t); getchar (); while (t --> 0) {gets (szStr); nMax =-1000; j = 1; I = 0; while (szStr [I]! = '\ 0') {a [j ++] = szStr [I]-65-13; I ++;} n = j-1; B [0] =-1000; nMax =-1000; for (I = 1; I <= n; ++ I) {B [I] = max (a [I], B [I-1] + a [I]); if (nMax <B [I]) {nMax = B [I] ;}} printf ("% d \ n", nMax);} return 0 ;}

 

/* Shard dynamic planning solution: However, the time and space complexity are optimized. In terms of time, it is only a traversal. In terms of space, strings are not used for saving because it is input and processed. Time: 7 ms memory: 196 K running */# include <stdio. h> # define max (x, y) (x)> (y )? (X): (y) int B [1004]; int main (void) {int I = 0; int j = 0; int t = 0; int n = 0; int nMax =-1000; int nTemp = 0; char chCur = '\ 0'; scanf ("% d", & t); getchar (); while (t --> 0) {nMax =-1000; B [0] =-1000; I = 1; while (chCur = getchar ())! = '\ N') {nTemp = chCur-65-13; B [I] = max (nTemp, B [I-1] + nTemp ); if (nMax <B [I]) {nMax = B [I];} I ++;} printf ("% d \ n", nMax);} return 0 ;}