Problem Description In college, a student may take several courses. for each course i, he earns a certain credit (ci), and a mark ranging from A to F, which is comparable to a score (si), according to the following conversion tableThe GPA is the weig…

GPA Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4802 Description In college, a student may take several courses. for each course i, he earns a certain credit (ci), and a mark ranging from A to F, which is c…

题意: 计算GPA,输入一个数字和一个字符串,用 数字×字符串对应的数值 思路: 用map对应数值,要注意的是字符串为P或者N的时候,不计入结果 代码: #include<iostream> #include<string> #include<cstring> #include<cstdio> #include<map> using namespace std; map<string, double> mp; int main() {…

Problem Description 每学期的期末,大家都会忙于计算自己的平均成绩,这个成绩对于评奖学金是直接有关的.国外大学都是计算GPA(grade point average) 又称GPR(grade point ratio),即成绩点数与学分的加权平均值来代表一个学生的成绩的.那么如何来计算GPA呢? 一般大学采用之计分法 A90 - 100 4 点 B80 - 89 3 点 C70 - 79 2 点 D60 - 69 1 点 E0 - 59 0 点 例如:某位学生修习三门课,其课目.…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

[目标] 剑灵GPA [思路] 1 2 绘制角色DrawCall body 5526面片 2.1[第一个DrawCall]63 RT 这个DrawCall PS VS 参数列表 VS // // Generated by Microsoft (R) HLSL Shader Compiler 9.29.952.3111 // // Parameters: // //   float4x3 BoneMatrices[75]; //   float4 CameraPosition; //   floa…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…