示例#1
0
文件: ariprog.c 项目: houjun/acm_icpc
int main()
{

	FILE *fin=fopen("pprime.in","r");
	FILE *fout=fopen("pprime.out","w");   
	int a,b,i,num;

	clock_t begin, end;
	double  cost;
	begin = clock();

	fscanf(fin,"%d %d",&a,&b);

	if((b-a)>500000)
		num=prime1(pnum,a,b);
	else{
		num=prime2(pnum,a,b);
		start=0;
	}

	for (i = start; i < num ; i++){
		if(pnum[i]>=a)
			if(ispal(pnum[i]))
				fprintf(fout,"%d\n",pnum[i]);		
	}

	end = clock();
	cost = (double)(end - begin) / CLOCKS_PER_SEC;
	printf("%lf seconds\n", cost);


	return 0;
}
示例#2
0
文件: 10848.c 项目: DavidToca/acm
int main()
{
	int p[8], i, j;

	while (gets(buf)) {
		memset(p, 0, sizeof(p));
		p[1] = check1();
		if (p[1]) {
			p[2] = ispal(s2);
			p[3] = check3();
			p[4] = check4();
			p[5] = check5();
			p[6] = check6();
			p[7] = check7();
		}

		for (i = 1, j = 1; i <= 7; i++) {
			putchar(p[i] ? 'T' : 'F');
			j &= p[i];
		}
		printf(" The solution is %s\n", j ? "accepted" : "not accepted");
	}

	return 0;
}
示例#3
0
main () {
    FILE *fin  = fopen ("dualpal.in", "r");
    FILE *fout = fopen ("dualpal.out", "w");
    int an, as;
	int i,j,t,ok;
	char s[33];
	
    fscanf (fin, "%d %d", &an, &as);
    
	t=0;
	for(i=as+1;;i++){
		ok=0;
		for(j=2;j<=10;j++){
			rnb(s,i,j);
			if(ispal(s)){
				if(++ok>1)break;
			}
		}
		if(ok>1){
			if(++t>an)break;
			fprintf (fout, "%d\n", i);
		}
	}

    exit (0);
}
示例#4
0
文件: dualpal.c 项目: euma/usaco
int main(void) {
    char t[20];
    int i, base;
    int n,s;
    FILE *fin, *fout;

    fin = fopen("dualpal.in", "r");
    fout = fopen("dualpal.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d %d", &n, &s);
    int get=0;
    for(i=s+1;get<n; i++) {
	int find=0;
	for (base=2;base<=10;base++) {
	    numbconv(t, i, base);
	    if(ispal(t)) {
		//printf("%s<%d> ", t, base);
		find++;
	    }
	    if (find > 1) {
	       	//printf("(%d)\n",i);
		get++;
	       	fprintf(fout,"%d\n",i);
		break;
	    }
	}
    }
    exit(0);
    return 0;
}
示例#5
0
文件: calfflac.c 项目: nanAdair/USACO
int main()
{
    FILE *fin = fopen("calfflac.in", "r");
    FILE *fout = fopen("calfflac.out", "w");
    char str[20001];
    char back[20001];
    int index[20001];
    int  i, j, k, smallindex, largeindex;
    i = j = k = 0;
    
    while(fscanf(fin, "%c", &back[j++]) > 0){
        if(back[j - 1] >= 'A' && back[j - 1] <= 'Z'){
            index[i] = j -1;
            str[i] = back[j - 1] - 'A' + 'a';
            i++;
        }
        else if(back[j - 1] >= 'a' && back[j - 1] <= 'z'){
            index[i] = j - 1;
            str[i] = back[j - 1];
            i++;
        }
    }
    str[i] = '\0';
    
    int max = 0;
    for(i = 0; i < j; i++){
        int temp;
        temp = ispal(str, i);
        if(temp > max){
            max = temp;
            if(max % 2 == 1){
                smallindex = i - max/2;
                largeindex = i + max/2;
            }
            else {
                smallindex = i - max/2 + 1;
                largeindex = i + max/2;
            }
        }
    }
    
    char re[2001];
    for(i = index[smallindex]; i <= index[largeindex]; i++)
        re[i-index[smallindex]] = back[i];
    re[i-index[smallindex]] = '\0';
    fprintf(fout, "%d\n%s\n", max, re);
    
    fclose(fin);
    fclose(fout);
}
示例#6
0
 int minCut(string s) {
     queue<int> node;
     node.push(0);
     int cuts = 0;
     //bfs
     vector<bool> visited(s.size(), false);
     while (true){
         queue<int> tmp;
         while (!node.empty()){
             int cur = node.front();
             node.pop();
             for (int i = s.size() - 1; i >= cur; i--){
                 if (visited[i]==false&&ispal(s, cur, i)){
                     if (i == s.size() - 1)
                         return cuts;
                     tmp.push(i+1);
                 }
             }
             visited[cur] = true;
         }
         cuts++;
         node = tmp;
     }
 }
示例#7
0
文件: dualpal.c 项目: chromano/USACO
void solve(int n, int s) {
    char buff[17];
    int i, base, npal;
    FILE *out;

    out = fopen("dualpal.out", "w");
    while (n) {
        npal = 0;
        s++;
        for (npal = 0, base = 2; base <= 10; base++) {
            convert(s, base, buff);
            if (ispal(buff))
                npal++;
            if (npal >= 2) {
                break;
            }
        }
        if (npal >= 2) {
            fprintf(out, "%d\n", s);
            n--;
        }
    }
    fclose(out);
}
示例#8
0
int main(){ 
    char s[100],ans[100],ans2[100];
    int i,k=0,r;
    scanf("%s",s);
   
    if(ispal(s)){
        for(i=0;i<strlen(s)/2;i++){
            ans[i]=s[i];
        }
        ans[i]=s[i];i++;
        for(;i<strlen(s)+1;i++){
            ans[i]=s[i-1];
        }
        ans[i]='\0';
    }
    else{        
        k=strlen(s);
        for(i=0;i<k;i++){
            if(s[i]!=s[k-i-1]){               
                if(s[i+1]==s[k-i-1]){
                    for(r=0;r<k-i;r++){
                        ans2[r]=s[r];
                    }
                    ans2[r++]=s[i];                    
                    ans2[k+1]='\0';                     
                    for(;r<=k;r++){
                        ans2[r]=s[r-1];
                    }                                        
                    break;
                }
                else if(s[i]==s[k-i-2]){    
                            
                    for(r=0;r<i;r++){
                        ans2[r]=s[r];
                    }
                    ans2[r]=s[k-i-1];r++;
                    for(;i<=k;i++,r++){
                        ans2[r]=s[i];
                    }
                   
                    break;
                }
                else{
                    ans2[0] = 'N';
                    ans2[1] = 'A';
                    ans2[2] = '\0';
                    break;
                }
            }
        }    
        for(i=0;i<k;i++){
            if(s[i]!=s[k-i-1]){               
                if(s[i]==s[k-i-2]){    
                            
                    for(r=0;r<i;r++){
                        ans[r]=s[r];
                    }
                    ans[r]=s[k-i-1];r++;
                    for(;i<=k;i++,r++){
                        ans[r]=s[i];
                    }
                    break;
                }
                else if(s[i+1]==s[k-i-1]){
                    for(r=0;r<k-i;r++){
                        ans[r]=s[r];
                    }
                    ans[r++]=s[i];                    
                    ans[k+1]='\0';                     
                    for(;r<=k;r++){
                        ans[r]=s[r-1];
                    }                                        
                    break;
                }
                else{
                    ans[0] = 'N';
                    ans[1] = 'A';
                    ans[2] = '\0';
                    break;
                }
            }
        }
        if(ispal(ans2)){
            for(i=0;i<=strlen(ans2);i++){
                ans[i]=ans2[i];
            }
        }
    }
    if(ispal(ans)==0){
        ans[0] = 'N';
        ans[1] = 'A';
        ans[2] = '\0';                
    }
    
    printf("%s",ans);
    return 0;    
}