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Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16488   Accepted: 11453 Description A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will…
quicksum Given a string of digits, find the minimum number of additions required for the string to equal some target number. Each addition is the equivalent of inserting a plus sign somewhere into the string of digits. After all plus signs are insert…
Quicksum 时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte总提交:615            测试通过:256 描述 A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the checksum will also change…
Quicksum Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 102   Accepted Submission(s) : 33 Problem Description A checksum is an algorithm that scans a packet of data and returns a single number.…
POJ3094 Quicksum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18517   Accepted: 12712 Description A checksum is an algorithm that scans a packet of data and returns a single number. The idea is that if the packet is changed, the check…
注意: for (int i = 1; i <= aaa.length(); i++) 其中是“ i <= ",注意等号. 原题: 2520.   Quicksum Time Limit: 0.5 Seconds   Memory Limit: 65536KTotal Runs: 2964   Accepted Runs: 1970 A checksum is an algorithm that scans a packet of data and returns a single…
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit cid=1006#status//H/0" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" style="font-family:Verdana,Arial,sans-serif; f…
Quicksum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3516 Accepted Submission(s): 2579 Problem Description A checksum is an algorithm that scans a packet of data and returns a single number. T…
http://poj.org/problem?id=3094 #include<iostream> #include <string> using namespace std; int main() { string str; getline(cin,str); while(str != "#") { ; ;i < str.length(); i++) { ; if(str[i] >= 'A' && str[i] <= 'Z')…
#include <stdio.h> #include <string.h> ]; int main() { ; int i,len; while(gets(word)) { sum = ; ] == '#') break; len = strlen(word); ; i < len; ++i) { if(word[i] != ' ') sum += (i+) * (word[i]-); } printf("%d\n",sum); } ; }…