JACK_JYH的博客

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int magic[105];int attack[105];int dp[105][105];int main(){ int n,t,q; while(scanf("%d%d%d",&n

import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n; char ch[]; while(in.hasNext())

import java.io.*;import java.util.*;public class Main{ public static void main(String args[]) { Scanner in=new Scanner(System.in); int T;

题目大意：你是一个战士现在面对，一群狼，每只狼都有一定的主动攻击力和附带攻击力。你杀死一只狼。你会受到这只狼的（主动攻击力+旁边两只狼的附带攻击力）这么多伤害~现在问你如何选择杀狼的顺序使的杀完所有狼时，自己受到的伤害最小。（提醒，狼杀死后就消失，身边原本相隔的两只狼会变成相邻，而且不需要考虑狼围城环这种情况）输入要求：总共T组数据，每组N只狼，按顺序输入全部狼的主动攻击力和然后再按顺序输入全部狼的

题目大意：ZZZ在上课，这堂课N分钟，他必须睡M分钟，如果他听课至少听L分钟的课，每分钟的课都有对应的分数，求如何使他在睡满M分钟的情况下，得分最高。输入要求：N分钟的课堂，至少睡M分钟，每次最少听L分钟（以EOF结尾）输出要求：最高得分样例数据分析： 输入10 3 31 2 3 4 5 6 7 8 9 10 前3分钟睡觉：4+5+6+7+8+9+10=49；输出49

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int a[25][1005];int dp[25][1005];int main(){ int C; scanf("%d",&C); while(C--) {

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dp[1001];int volume[1001],value[1001];int main(){ int t; scanf("%d",&t); while(t--)

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;//#include<algorithm>using namespace std;为了调用库函数maxint dp[1005][1005];int l[1005],r[1005];int main(){

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;//#include<algorithm>using namespace std;为了调用库函数max__int64 dp[100005]; //因为高已经很大了乘上底会更大，所以用__int64__int64 h[100005],l[10000

#include<stdio.h>int main(){ int t,n,a,i,j,start,end,temp; start=1,end=1; long max,sum; scanf("%d",&t); for(i=1;i<=t;i++) { sum=0; temp=1; max=-1001;

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;//#include<algorithm>using namespace std;为了调用库函数maxint dp[10005];int main(){ int a[10005]; int start[10005],end[1

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;//#include<algorithm> using namespace std;这两句是因为下面要调用库函数中的max函数double dp[10005];//储存的是不被抓到的概率，最多一百家银行，每家最多抢100，所以取10005stru

题目大意：有 n 个人，每个人有一个diaosi[i]值，如果第 i 个人排在第 k 位置，则他的愤怒值就为diaosi[i]*（k-1）；等待过程中有一个黑屋子，可以把人暂时放到黑屋子里。使总的愤怒值最小；输入要求：T组数据，n个人，每个人的愤怒值输出要求：Case #2: 最小愤怒值思路引导：这是一道区间dp题。我们用dp[i][j]表示第i个人到第j个人unhappiness总和的最小值，因此

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;//#include<algorithm>using namespace std;为了调用库函数maxint dp[3000005];//最多三十张每张最多1000，因为小数点后两位再乘100int main(){ double Q;

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int a[1005];int dp[1005];int main(){ int N; int i,j; while(scanf("%d",&N)) {

package Main;import java.io.*;import java.util.*;public class Main{ public static void main(String args[]) { Scanner in=new Scanner(System.in);

题目大意：每次输入两串字符串，把第一串变成第二串，但是你每次只能把字符串中一段连续的子串变成一同一种字符串，问你总共需要几次才能完成？输入要求：两串字符，字符串最长为100。输出要求：需要改变的次数。样例解释： 开始zzzzzfzzzzz 第一次：aaaaaaaaaaa 第二次：abbbbbbbbba 第三次：abcccccccba 第四次：abcdddddcba 第五次：abcdeee

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;bool dp[205][205];char str1[205],str2[205],str3[410];int main(){ int T; scanf("%d",&T); str

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int number[105],price[105];int dp[105];int main(){ int T; scanf("%d",&T); while(T--) {

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dp[200005];int main(){ int n; while(scanf("%d",&n)!=EOF) { int i; int a,

#include<iostream> #include<cstdio>#include<cstring>#include<algorithm> #include<stack>using namespace std;const int INF=0x3f3f3f3f;struct node1{ string name; int deadline; int cost;

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dp[1005][1005];char str1[1005],str2[1005];int map[5][5]={{5,-1,-2,-1,-3}, {-1,5,-3,-2

单调队列#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int n,m;int i,j,k;int p[105],h[105],c[105];int dp[2500];void MPqueue(int dp[], int SumValue, int v

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;char map[1005][1005];int dp[1005][1005];int l[1005];int r[1005];int Max;int n,m;void DP(){ int

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dp[105][2];char str[105];int main(){ int num; int i; scanf("%d\n",&num); while(num-

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int a[105], b[105];int dp[105][105];int main(){ int N, M, K, S, i, j, k; while(scanf("%d%d%d%d",

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;double dp[10005];struct node{ int money; double pro;}school[10005];int main(){ int n,m;

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dp[105][105];int a[105][105];int main(){ int C; scanf("%d",&C); while(C--) {

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int dp[500000];int value[1005];int number[1005];int MidSum;int Sum;int i;void Complete()//完全背包{

#pragma warning(disable:4996)#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include<algorithm>using namespace std;const int INF = 0x3f3f3f3f;int route[15][15];int dp[1 <

未完待续#pragma warning(disable:4996)#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include<algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int MAX = (1 << 16)

题目意思：Ignatius同学，快到期末了结果还有好多课程没有完成，每门课作业都有对应的截至日期，和完成它所需的时间，超过截至日期后，没多一天扣1分，现在文他该如何安排写作业的顺序使得被扣的分最少？输入要求：T组数据，N门课，每门课的截至日期和完成所需的时间输出要求：被扣的分，和写作业的顺序。（如果有不同顺序扣分相同的出现，输出字典序）题目提示：1.因为最多只有15门课，所以1<<15在一个可以接受

中文不解释~#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;int a[20],p[20];int dp[(1<<20)][20];int main()

#include<stdio.h>#include<string.h>#define max(a,b)(a)>(b)?(a):(b)#define INF 0x3f3f3f3fint g[155][155];int dp[1<<16][16];int num[22];int earn[22];int cost[22];int main(){ int t,i,j,k,mon

#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<algorithm>#include<iostream>#include<time.h>using namespace std;int dp[105][105];int mark[105];int pat

#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<algorithm>#include<iostream>#include<time.h>using namespace std;int dp[35][205];int pos[205];int f[205

public class MainTest { public static void main(String args[]) {//String[] args也行 int n=args.length; if(n==0) System.out.println("no parament !"); else {

import java.util.Scanner;public class Main{ public static void main(String[] args) { double x1,x2,y1,y2; Scanner in=new Scanner(System.in); while(in.hasNext()) {

import java.math.BigInteger;import java.util.Scanner;public class Main { public static void main(String[] args) { int T; Scanner in=new Scanner(System.in); T=in.nextInt();

import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner in=new Scanner(System.in); int a; int b; while(in.hasNextInt())